Entire functions of finite order with prescribed zeros

I'm trying to solve the following the problem.

Let $f$ be an entire function of finite order that has roots at all points $z$ such that $e^{e^z}=1$. Prove that $f\equiv 0$.

My idea is argue by contradiction, suppose $f$ is not a constant and consider the function $h(z) = \frac{f(z)}{e^{e^z}-1}$. I want to show that $h$ is a constant, then $f$ is a multiple of $e^{e^z}-1$ and has infinite order. But I'm stuck here now and actually I don't know if this works. There might be such an $f$ with infinite order but not a multiple of $e^{e^z}-1$.

Do we have some results about two functions share the same (or partial) zeros?

Solution 1:

$e^{e^z}=1$ means $e^z=2k\pi i$ which means $z=\log (2|k|\pi)\pm \pi i/2+2m\pi i, k,m \in \mathbb Z, k \ne 0$ where the sign is plus if $k>0$ and minus if $k<0$

In particular, if we fix $R>0$ large one can choose $m=0, 1\le k \le \frac{e^{R-\pi/2}}{2\pi}$ and get a root as above so we get at least $Ce^R$ roots; however a function of order $\rho$ has at most $C_{\epsilon}R^{\rho+\epsilon}$ roots of absolute value at most $R$ so contradiction!

Solution 2:

$e^{e^{z}}=1$ iff $z=\ln(2k\pi)+(2l\pi+\frac{\pi}{2})i, k\in \mathbb{N}^{*},l\in \mathbb{Z}$.

We will use Theorem 2.1 (ii) in Ch5 of Stein and Shakarchi's complex analysis.

Suppose $f$ is not the zero function and the order of $f$ is $\rho$. Then $\Sigma_{k=1}^{\infty}\frac{1}{{|\ln(2k\pi)+\frac{\pi}{2}|}^{s}}$ converges for all $s>\rho$. But we know this series diverges for all $s>0$. Hence $f=0$.