$z_0$ 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 $f$.

Under aforementioned conditions the function $f(z)$ may be expanded about $z = z_0$ 
up to a certain radius (radius of convergence) namely
\begin{equation*}
f(z) = \sum_{k= -\infty}^{+ \infty} a_k (z-z_0)^k = 
\sum_{k= 1}^{+\infty} \frac{a_{-k}}{ (z-z_0)^k} + \sum_{k= 0}^{+ \infty} a_k (z-z_0)^k = f_0(z) + f_1(z);
\end{equation*}
here we note the principal part $f_0(z)$ and the non-principal holomorphic part $f_1(z)$.
Because the isolated singularity of $f(z)$ at $z=z_0$ is not removable, it means that we have either a pole of order $n \in \mathbb{N}$ or an essential singularity.
In case of a pole at least $a_{-n} \ne 0$, in case of an essential singularity infinitely many coefficients $a_{-k},\ k \in \mathbb{N}$ are non-zero - 
in both cases the principal part $f_0(z)$ is not trivially equivalent to zero.
We may inspect $g(z) = e^{f(z)}$ and its expression about $z=z_0$ namely
\begin{equation*}
g(z) = e^{f(z)} %= e^{ f_0(z) + f_1(z)} = e^{f_0(z)} \cdot e^{f_1(z)} = 
= \sum_{j=0}^{+\infty} \frac{(f_0(z))^j}{j!}  \cdot e^{f_1(z)} = 
\sum_{j=0}^{+\infty} \frac{1}{j!} \left( \sum_{k= 1}^{+\infty} \frac{a_{-k}}{ (z-z_0)^k} \right)^j  \cdot e^{f_1(z)} = g_0(z) \cdot g_1(z);
\end{equation*}
here we note the non-trivial multiplicand $g_0(z)$ and the leftover $g_1(z) = e^{f_1(z)}$ 
is a holomorphic non-vanishing function in a neighborhood of $z=z_0$. 
However it is clear that the principal part of $g_0(z)$ exhibits an infinitude of summands in the obtained expression - 
it is equivalent then to $g(z)$ having an essential singularity at $z=z_0$.
