Question regarding essential singular point of a function 
How to prove the function $$ f(z)=\exp\Big(\frac{z}{1-\cos z}\Big)$$ has an essential singularity at $z=0$ ?

It's actually hard to express the Laurent series of $f(z)$ around $0$, because the power $\frac{z}{1-\cos z}$ itself is already in the series form (since $\cos z$ appears there and it has the series expansion) and $e^{z/(1-\cos z)}$ has again a series form.  
Edit 1: I already see this but it does not give information about the Laurent expension of $f(z)$ 
Edit 2: How to Proceed or can anyone explain the limit of  $e^{z/(1-\cos z)}$ at $0$ does not exist?
 A: It's sufficient to prove that the limit $$a = \lim_{z \to 0} \exp \left( \frac{z}{1-\cos z} \right)$$ does not exist, for we have the following trichotomy:


*

*if $z$ is a removable singularity, the limit exists and $a \in \mathbb{C},$

*if $z$ is a pole, then $a = \infty$ (the complex infinity),

*if $z$ is an essential singularity, the limit does not exist.

A: $\exp \left( \frac{z}{1-\cos \left( z \right)} \right)=\exp \left( \frac{2}{z}+\frac{z}{6}+\frac{{{z}^{2}}}{120}+... \right)=\exp \left( \frac{2}{z}+O\left( z \right) \right)=\exp \left( O\left( z \right) \right)\sum\limits_{n=0}^{\infty }{\frac{{{2}^{n}}}{n!{{z}^{n}}}}$
...or am i missing something?
Edit:  if that's not enough, note:
$$\begin{align}
  & \frac{z}{1-\cos \left( z \right)}=\frac{z}{2}{{\csc }^{2}}\left( \frac{1}{2}z \right)=\frac{z}{2}{{\left( \sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}2\left( {{2}^{2n-1}}-1 \right){{B}_{2n}}}{\left( 2n \right)!}}{{\left( \frac{z}{2} \right)}^{2n-1}} \right)}^{2}} \\ 
 & =\frac{z}{2}{{\left( \frac{2}{z}+\frac{z}{12}+... \right)}^{2}}=\frac{2}{z}{{\left( 1+\frac{{{z}^{2}}}{24}+... \right)}^{2}}=\frac{2}{z}+O\left( z \right) \\ 
\end{align}$$
A: We wish to show that the function's Laurent series has infinitely many negative terms. As you said, it may be difficult to express the entire Laurent series, but there is a formula for each of the Laurent coefficients:
$$a_n = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z^{n + 1}}dz$$
where $\gamma$ is a curve around $0$ within an annulus on which $f$ is holomorphic. Perhaps you could try to show that there is no $n < 0$ such that for all $k < n$, $a_k = 0$.
