Let $z_0$ be a pole of function $f(z)$. Prove that $z_0$ is an essential singularity of $e^{f(z)}$.
I already know that $e^{f(z)}$ when $z\to z_0$ could be unbounded so $z_0$ should be a pole or singularity. But when $z_0$ is a pole I can't find contradiction.
Namely I want to find a $z'$ in the neighbor of $z_0$ which satisfies $f(z')$ is pure imaginary then $|e^{f(z')}|=1$, contradicts with $e^{f(z)}$ takes $z_0$'s neighbor to $\infty$'s neighbor. Is my thought correct? Thanks for any help.
An idea: if $\;z_0\;$ is a pole of $\;f\;$ , then in some neighborhood of it we have a Laurent series for the function:
$$f(z)=\sum_{n=-k}^\infty a_n(z-z_0)^n\;,\;\;a_{-k}\neq0\implies\text{using the series for the exponential around}\;\;z_0:$$
$$e^{f(z)}=e^{a_{-k}(z-z_0)^{-k}+\ldots}=1+a_{-k}(z-z_0)^{-k}+\frac{\left(a_{-k}(z-z_0)^{-k}\right)^2}2+\frac{\left(a_{-k}(z-z_0)^{-k}\right)^3}6+\ldots$$
and we get infinite negative powers in the above development of $\;e^{f(z)}\;$ as powers of $\;z-z_0\implies z_0\;$ is an essential singularity.