Find the Laurent series expansion of $\frac{1}{z(e^z-1)}$ and show that it has a pole of second order at $z = 0$. I am following an example solution for finding the Laurent expansion for $\frac{1}{z(e^z-1)}$, $0 <|z|< 2\pi$. The solution is given as:
\begin{align*}\frac{1}{z(e^z-1)} &= \frac{1}{z(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+ \ldots-1)} = \frac{1}{z(z+\frac{z^2}{2!}+\frac{z^3}{3!}+ \ldots)} \\ &= \frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)} = \frac{1}{z^2}[1+(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)]^{-1} \\ &= \frac{1}{z^2}[1-(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)+(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)^2 - \ldots] \\ &= \frac{1}{z^2}[1-\frac{z}{2!} + z^2(-\frac{1}{3!}+ \frac{1}{4}) - z^3(-\frac{1}{4!}+ \frac{1}{3!} -\frac{1}{8}) + \ldots] \\ &= \frac{1}{z^2} - \frac{1}{2z} + \frac{1}{12} + \frac{1}{360z^2} + \ldots\end{align*}
I am able to understand this except the expansion of $[1+(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)]^{-1}$. My question is how did we ensure that $|(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)| <1$ so that the binomial formula $(1+z)^{-1} $ is applicable?

There's another way, using the Bernoulli  numbers
$$
\frac{z}{e^z-1}=\sum_{n\geq 0}\frac{B_n}{n!}z^n
$$
Then,
$$
\frac{1}{z(e^z-1)}=\frac{1}{z^2}\frac{z}{e^z-1}=\sum_{n\geq 0}\frac{B_n}{n!}z^{n-2}
$$
Hope it was useful!
