JavaScript (ES6), 58 54 bytes

a=>a.map(c=>a<(a=c)?`Excelsior${s+='!'}

`:s='').join``

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Commented

a =>                           // a = input array, also used to store the previous value

  a.map(c =>                   // for each value c in a:

    a <                        //   compare the previous value

    (a = c)                    //   with the current one; update a to c

                               //   this test is always falsy on the 1st iteration

    ?                          //   if a is less than c:

      `Excelsior${s += '!'}n` //     add a '!' to s and yield 'Excelsior' + s + linefeed

    :                          //   else:

      s = ''                   //     reset s to an empty string and yield an empty string

  ).join``                     // end of map(); join everything

Why re-using a[ ] to store the previous value is safe

There are three possible cases:

• If $a[text{ }]$ is empty, the callback function of .map() is not invoked at all and we just get an empty array, yielding an empty string.


• If $a[text{ }]$ contains exactly one element $x$, it is coerced to that element during the first (and unique) test a < (a = c). So, we're testing $x < x$, which is falsy. We get an array containing an empty string, yielding again an empty string.


• If $a[text{ }]$ contains several elements, it is coerced to NaN during the first test a < (a = c). Therefore, the result is falsy and what's executed is the initialization of $s$ to an empty string -- which is what we want. The first meaningful comparison occurs at the 2nd iteration.

Python 2, 84 83 81 70 68 bytes

a=n=''

for b in input():

 n+='!';n*=a<b;a=b

 if n:print'Excelsior'+n

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_{-2 bytes, thanks to ASCII-only}

05AB1E, 26 24 23 bytes

ü‹γvyOE.•1Š¥èò²•™N'!׫,

-2 bytes thanks to @Kroppeb.

Try it online or verify all test cases.

Explanation:

ü                        # Loop over the (implicit) input as pairs

 ‹                       #  And check for each pair [a,b] if a<b is truthy

                         #   i.e. [1,2,1,3,4,1,5,7,20,25,3,17]

                         #   → [1,0,1,1,0,1,1,1,1,0,1]

  γ                      # Split it into chunks of equal elements

                         #  i.e. [1,0,1,1,0,1,1,1,1,0,1]

                         #   → [[1],[0],[1,1],[0],[1,1,1,1],[0],[1]]

   vy                    # Foreach `y` over them

     O                   #  Take the sum of that inner list

                         #   i.e. [1,1,1,1] → 4

                         #   i.e. [0] → 0

      E                  #  Inner loop `N` in the range [1, length]:

       .•1Š¥èò²•         #   Push string "excelsior"

                ™        #   Titlecase it: "Excelsior"

                 N'!׫  '#   Append `N` amount of "!"

                         #    i.e. N=3 → "Excelsior!!!"

                      ,  #   Output with a trailing newline

See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•1Š¥èò²• is "excelsior".

Perl 6, 60 58 57 bytes

-1 byte thanks to nwellnhof

{"Excelsior"X~("!"Xx grep +*,[[&(-+^*×*)]] .skip Z>$_)}

Try it online!

Anonymous code block that returns a list of Excelsiors!

Java-8 118 113 Bytes

n->{String e="";for(int i=0;i<n.length-1;)System.out.print(""==(n[i+1]>n[i++]?e+="!":(e=""))?e:"Excelsior"+e+"n");}

Easy to read :

private static void lee(int num) {

    String exclamation = "";

    for (int i = 0; i < num.length - 1;) {

        exclamation = num[i + 1] > num[i++] ? exclamation += "!" : "";

        System.out.print("".equals(exclamation) ? "" : "Excelsior" + exclamation + "n");

    }

}

R, 86 bytes

Half of this answer is @Giuseppe's. RIP Stan Lee.

function(a)for(i in diff(a))"if"(i>0,cat("Excelsior",rep("!",F<-F+1),"

",sep=""),F<-0)

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05AB1E, 20 19 bytes

ü‹0¡€ƶ˜ε'!×”¸Îsiorÿ

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Explanation

ü‹                    # pair-wise comparison, less-than

  0¡                  # split at zeroes

    €ƶ                # lift each, multiplying by its 1-based index

      ˜               # flatten

       ε              # apply to each

        '!×           # repeat "!" that many times

                  ÿ   # and interpolate it at the end of

           ”¸Îsior    # the compressed word "Excel" followed by the string "sior"

C (gcc/clang), 106 99 97 bytes

f(a,n)int*a;{int r=0,s[n];for(memset(s,33,n);n-->1;)r*=*a<*++a&&printf("Excelsior%.*sn",++r,s);}

Thanks to gastropner for golfing 2 bytes.

Try it online here.

Ungolfed:

f(a, n) // function taking a pointer to the first integer and the length of the array

  int *a; { // a is of type pointer to int, n is of type int



    int r = 0, // length of the current run

        i = 0, // loop variable

        s[n];  // buffer for exclamation marks; we will never need more than n-1 of those (we are declaring an array of int, but really we will treat it as an array of char)



    for(memset(s, 33, n); // fill the buffer with n exclamation marks (ASCII code 33)

        n -- > 1; ) // loop over the array



        r *= *a < *(++ a) // if the current element is less than the next:

             && printf("Excelsior%.*sn", // print (on their own line) "Excelsior", followed by ...

                       ++ r, // ... r (incremented) of the ...

                       s) // ... n exclamation marks in the buffer s

             ; // else r is reset to 0



}

Japt -R, 25 22 bytes

ò¨ ËÅ£`Ex­lÐâ`ú'!Y+A

c

Try it

3 bytes saved thanks to Kamil

ò¨                      :Partition at items that are greater than or equal to the previous item

   Ë                    :Map

    Å                   :  Slice off the first element

     £                  :  Map each element at 0-based index Y

      `Ex­lÐâ`           :    Compressed string "Excelsior"

             ú'!        :    Right pad with exclamation marks

                Y+A     :     To length Y+10

c                       :Flatten

                        :Implicitly join with newlines & output

Common Lisp, 111 bytes

(setq i 0)(loop for(a b)on(read)do(incf i(if(and b(> b a))1(- i)))(format(> i 0)"Excelsior~v@{~a~:*~}~%"i #!))

Try it online!

Java 8, 106 bytes

n->{String s="",z=s;for(int i=0;i<n.length-1;)z+=n[i++]<n[i]?"Excelsior"+(s+="!")+"n":(s="")+s;return z;}

Try it online!

^{(those reassignments of s...yikes)}

Stax, 17 bytes

Θx7├╖&σ '@7g┼┘Ñ«═

Run and debug it

R, 111 bytes

function(a,r=rle(sign(diff(a))),v=r$l[r$v>0])write(paste0(rep("Excelsior",sum(v)),strrep("!",sequence(v))),1,1)

Try it online!

A far better R tribute can be found here -- I was too fixated on sequence and rle.

Jelly, 16 bytes

<Ɲṣ0ÄẎ”!ẋ“Ø6ḥ»;Ɱ

A monadic Link yielding a list of lists of characters.

Try it online! (footer joins with newlines)

How?

<Ɲṣ0ÄẎ”!ẋ“Ø6ḥ»;Ɱ - Link: list of integers     e.g. [1,1,4,2,1,1,3,4]

 Ɲ               - for each pair of integers:      [1,1] [1,4] [4,2] [2,1] [1,1] [1,3] [3,4]

<                -   less than?                    [  0,    1,    0,    0,    0,    1,    1]

  ṣ0             - split at zeros                  [,    [1],     ,   ,      [1,    1]]

    Ä            - cumulative sums                 [,    [1],     ,   ,      [1,    2]]

     Ẏ           - tighten                         [1,1,2]

      ”!         - literal '!' character           '!'

        ẋ        - repeat (vectorises)             [['!'],['!'],['!','!']]

         “Ø6ḥ»   - dictionary lookup               ['E','x','c','e','l','s','i','o','r']

               Ɱ - map with:

              ;  -   concatenate                   [['E','x','c','e','l','s','i','o','r','!'],['E','x','c','e','l','s','i','o','r','!'],['E','x','c','e','l','s','i','o','r','!','!']]

Perl 5 -n, 41 bytes

$_>$l&&$l?say$c.='!':($c=Excelsior);$l=$_

Try it online!

Takes its input on separate lines.

Japt, 22 bytes

ò¨ ®£`Ex­lÐâ`+'!pYÃÅÃc

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Explanation, with simplified example:

ò¨                       :Split whenever the sequence does not increase

                           e.g. [2,1,1,3] -> [[2],[1],[1,3]]

   ®               Ã     :For each sub-array:

    £            Ã       :  For each item in that sub-array:

     `Ex­lÐâ`             :    Compressed "Excelsior"

            +            :    Concat with

             '!pY        :    a number of "!" equal to the index

                               e.g. [1,3] -> ["Excelsior","Excelsior!"]

                  Å      :  Remove the first item of each sub-array

                            e.g. [[Excelsior],[Excelsior],[Excelsior,Excelsior!]]->[,,[Excelsior!]]

                    c    :Flatten

                           e.g. [,,[Excelsior!]] -> [Excelsior!]

Retina, 55 bytes

d+

*

L$rv`(_*,(?<!(?(1)1|2,)))+(_+)b

Excelsior$#1*!

Try it online! Link includes test cases. Explanation:

d+

*

Convert to unary.

rv`(_*,(?<!(?(1)1|2,)))+(_+)b

Process overlapping matches from right to left (although the matches are then listed from left to right). This means that we can match every number in a run, and the match extends to the start of the run. Each match is further constrained that each additional matched number must be less than the previously matched additional number, or the first number if no additional numbers have been matched yet.

L$...

Excelsior$#1*!

For each match, output Excelsior with the number of additional numbers in the run as desired.

Pyth, 32 bytes

j+L"Excelsior"*L!fT.u*hN<0Y.+Q0

Try it online here, or verify all the test cases at once here.

j+L"Excelsior"*L!fT.u*hN<0Y.+Q0   Implicit: Q=eval(input())

                            .+Q    Get forward difference between consecutive elements of Q

                    .u         0   Reduce the above, returning all steps, with current value N starting at 0, next element as Y, using:

                       hN            N+1

                      *              Multiplied by

                         <0Y         1 if 0<Y, 0 otherwise

                  fT               Filter to remove 0s

              *L!                 Repeat "!" each element number of times

 +L"Excelsior"                     Prepend "Excelsior" to each

j                                  Join on newlines, implicit print

Jelly, 18 bytes

<Ɲ‘×¥ḟ0”!ẋ“Ø6ḥ»;Ɱ

Try it online!

Output prettified over TIO.

Lua, 88 87 83 82 96 95 113 bytes

Thanks @Kevin Cruijssen for update adhering to spirit of original question.

s=io.read()n=9 e="Excelsior!"f=e

for c in s.gmatch(s,"%S+")do if n<c+0then print(e)e=e..'!'else e=f end n=c+0 end

Try it online!

C++ 14 (g++), 123 118 bytes

(auto a){for(int n=0,i=0;++i<a.size();)a[i]>a[i-1]?puts(&("Excelsior"+std::string(++n,33))[0]):n=0;}

Fortunately std::string has a constructor that repeats a char. Try it online here.

Thanks to gastropner for saving 5 bytes.

Ungolfed:

 (auto a) { // void lambda taking a std::array of integer



    for(int n = 0, // length of the current run

        i = 0; // loop variable

        ++ i < a.size(); ) // start with the second element and loop to the last

        a[i] > a[i - 1] // if the current element is greater than the previous ...

        ? puts( // ... print a new line:

               &("Excelsior" + // "Excelsior, followed by ...

                std::string(++ n, 33)) // ... the appropriate number of exclamation marks (33 is ASCII code for '!'); increment the run length

               [0]) // puts() takes a C string

        : n = 0; // else reset run length



}

C# (.NET Core), 115 107 105 bytes

a=>{var b="";for(int i=0;++i<a.Length;)if(a[i]>a[i-1])Console.WriteLine("Excelsior"+(b+="!"));else b="";}

Try it online!

-8 bytes: changed b to a string holding "!"s from an int counter
-2 bytes: set b+="!" as an inline function (thanks to Zac Faragher)

Uses an Action delegate to pull in the input and not require a return.

Ungolfed:

a => {

    var b = "";                         // initialize the '!' string (b)

    for(int i = 0; ++i < a.Length;)     // from index 1 until the end of a

        if(a[i] > a[i - 1])                 // if the current index is greater than the previous index

            Console.WriteLine("Excelsior" +     // on a new line, print "Excelsior"

                                    (b += "!"));    // add a "!" to b, and print the string

        else                                // if the current index is not greater than the previous index

            b = "";                             // reset b

}

PHP, 117 109 bytes

<?php do{$i=next($argv);if($p!==null&&$p<$i){$e.='!';echo "

Excelsior$e";}else$e='';$p=$i;}while($i!==false);

Try it online!

Java, 113 bytes

String i="";for(int a=0;a<s.length-1;a++){if(s[a+1]>s[a]){i+="!";System.out.println("Excelsior"+i);}else{i="";}}

VBA, 114 bytes

For i=0 To UBound(a)-LBound(a)-1 If a(i+1)>a(i)Then s=s&"!" Debug.Print("Excelsior"&s&"") Else s="" End If Next i

Python 3, 87 bytes

c='!'

for i in range(1,len(n)):

    if n[i]>n[i-1]:print('Excelsior'+c);c+='!'

    else:c='!'

Or 97 with the following:

c='!';n=input()

for i in range(1,len(n)):

    if n[i]>n[i-1]:print('Excelsior'+c);c+='!'

    else:c='!'

This assumes inputs will be in the format:

32105

12345

<null input>

1

1213415

333343

Japt, 25 bytes

ä< ®?`Ex­lÐâ`+'!p°T:T=0

f

Try it online!

J, 50 bytes

'Excelsior',"1'!'#"0~[:;@(([:<+/);._1)0,2</ ::0]

Try it online!

ungolfed

'Excelsior' ,"1 '!' #"0~ [: ;@(([: < +/);._1) 0 , 2 </ ::0 ]