How do I show that $\frac 1{e^z-1}$ has essential singularities (instead of say, poles) at $z=2n\pi i(n\in \mathbb Z)$?
I can't figure out how to show that the function does not go to infinity near $0$, or that it assumes every possible value near $0$. Exhibiting the laurent series around $0$ isn't general enough to show that essential singularities occur at all the stated points.
An idea: using, for example, l'Hospital, you could show that
$$\forall\,n\in\Bbb N\;,\;\;\lim_{z\to 2n\pi i}\frac{z-2n\pi i}{e^z-1}\stackrel{\text{l'H}}=\lim_{z\to 2n\pi i}\frac1{e^z}=1$$
so the singularity $\,z=2n\pi i\;$ is, in fact, a pole: a simple one with residue $\,1\,$
You don't!
The function in question has simple poles. The easiest way to see this is to note that $\exp(2n\pi i)=1$, so by continuity $e^z-1$ tends to zero as $z \rightarrow 2n\pi i$ and thus the reciprocal tends to infinity.
To show that the poles are simple, note that $e^z-1$ has a zero of degree $1$ at $2n\pi i$ so the reciprocal has a simple pole and so the residue is:
$$\frac{1}{\frac{d}{dz}(e^z-1)}=\frac{1}{e^{2n\pi i}}=1$$
