$z_0$ non-removable singularity of $f\Rightarrow z_0$ essential singularity of $\exp(f)$ 

Let $z_0$ be a non-removable isolated singularity of $f$. Show that $z_0$ is then an essential singularity of $\exp(f)$.


Hello, unfortunately I do not know how to proof that. To my opinion one has to consider two cases:


*

*$z_0$ is a pole of order $k$ of $f$.

*$z_0$ is an essential singularity of $f$.

The above proof is very nice, here I provide another idea. The real question is that if $f$ has a pole, then $e^f$ has an essential singularity.
Assume that the conclusion is false, then there exist an integer $k$ and an analytic function $g$ in some disc $|z|<\delta$, for example, $g(0)\neq0$, such that $e^f=z^kg$,
$$f'=\frac{(e^f)'}{e^f}=\frac{k}{z}+\frac{g'}{g}$$
integrating it on the contour $|z|=\frac{\delta}{2}$ we find $k=0$, since $g$ is analytic. Using the fact that $g(0)\neq0$ we can find another analytic function $h$ in the disc $|z|<\delta$(actually we should shrink this disc), such that $e^f=g=e^h$, so $f=h+2n\pi i$, $f$ has a removable singularity, contradiction!
