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, $\{a_{n}\}_{n} = \emptyset$.
Is there any way to get $$f(z) = ce^{p(z)}$$ with $c$ constant?
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.