pole becomes essential singularity when lifting by exponential 
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.
 A: 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.
A: I'm not sure if I understand DonAntonio's proof. It seems unclear to me why the coefficients cannot somehow cancel each other.
My proof is this, without loss of generality let $z_0=0$, let $f=g/z^n$ be a meromorphic function near $0$, where $g$ is holomorphic near $0$ and $g(0)\not=0$. If it is not true that $e^f$ has an essential singularity at $0$, then for some $m\in\mathbb{Z}^+$, we have  $$|z|^me^{g/z^m}\rightarrow 0$$ as $z\rightarrow 0$.
Since $$|z|^me^{g/z^m}=\exp\big(\frac{g}{z^m}+m\log|z|\big),$$
we see that $$\text{Re }\bigg(\frac{g}{z^m}+m\log|z|\bigg)\rightarrow -\infty$$ as $z\rightarrow 0$.
However, let $z=\lambda r$, where $r>0$ and $\lambda=g(0)^{1/m}$, and let $h(z)=g(z)/g(0)$, we see that $$\frac{\text{Re }h(\lambda r)}{r^m}+m\log r\rightarrow-\infty$$ as $r\rightarrow 0$.
However, since $h(0)=1$, we have $\text{Re }h(\lambda r)>\frac{1}{2}$ for all $r$ small enough, therefore, the above limit doesn't hold. This shows that $0$ is an essential singularity of $e^f$.
