Find the Laurent expansion of the following around $z_0 = 0$
\begin{align*} f(z) &= \frac{1}{z(e^z - 1)} \\ \end{align*}
The answer is
\begin{align*} f(z) &= \frac{1}{z^2} - \frac{1}{2z} + \frac{1}{12} - \frac{z^2}{720} + \frac{z^4}{30240} + \cdots \end{align*}
But I don't see how to find that by hand. I tried:
\begin{align*} f(z) &= -\frac{1}{z} \cdot \frac{1}{(1 - e^z)} = -\frac{1}{z} \cdot \sum\limits_{n=0}^\infty e^{zn}\\ \end{align*}
or
\begin{align*} f(z) &= \frac{1}{z} \cdot \frac{1}{e^z - 1} = \frac{1}{z} \cdot \frac{1}{\left( \sum\limits_{n=0}^\infty \frac{z^n}{n!} \right) - 1} \\ &= \frac{1}{\sum\limits_{n=1}^\infty \frac{z^{n+1}}{n!}} \\ \end{align*}
Neither of the two approaches I tried seem to get me to the solution. Any ideas?