Krdytkyu

Suppose $f$ is an entire and never vanishes, and that none of higher derivatives of $f$ ever vanish. Prove that if $f$ is also of finite order, then $f(z) = e^{az+b}$ for some constants $a,b$.

I have an idea that doesn't seem to work on two points. I want to know if there is any way to take advantage of this idea or if there is some other way.

My approach. By Hadamard's Formula,
$$f(z) = e^{p(z)}z^{m}prod_{1}^{infty}E_{k}(z/a_{n})$$
where $p$ is a polynomial of degree $leq k$, $m$ is the order of the zero of $f$ at $z=0$ and ${a_{n}}_{n}$ denotes the non-zero zeros of $f$.
But in this case, ${a_{n}}_{n} = emptyset$.

1. Is there any way to get
   $$f(z) = ce^{p(z)}$$
   with $c$ constant?




2. If yes, we have $f'(z) = cp'(z)e^{p(z)}$. My idea is to use the hypothesis about the higher derivatives to conclude that $p'(z)$ is constant. But the higher derivatives may have zeros.

$f$ has no zeros, therefore $m=0$ and – as you noticed – the set ${ a_n }$ of non-zero zeros is empty. So
$ f(z) = e^{p(z)} $ for some polynomial $p$, and $f'(z) = p'(z) e^{p(z)} $.

If $f'$ never vanishes then $p'$ is a polynomial without zeros, and therefore
constant. It follows that $p$ is a polynomial of degree at most one.