# Residue of $(1-e^{-z})^{-n}$ at 0

Similar to Evaluating the residue of $$(1 - e^{-z})^n$$ at $$z = 0$$ with $$n \in \mathbb{Z}$$, but this question is unanswered.

What is the residue at $$0$$ of $$(1-e^{-z})^{-n}$$?

For $$n=1$$, we can just differentiate $$1-e^{-z}$$ and so $$\text{res}_0f=1/1=1$$, but for $$n>1$$ I'm stuck.

$$\operatorname*{Res}_{z=0}\frac{1}{(1-e^{-z})^n}=\frac{1}{2\pi i}\oint\frac{dz}{(1-e^{-z})^n}\stackrel{z\mapsto \log(1+t)}{=}\frac{1}{2\pi i}\oint\frac{dt}{(1+t)\left(\frac{t}{1+t}\right)^n}$$ equals $$\frac{1}{2\pi i}\oint\frac{(1+t)^{n-1}}{t^n}\,dt = \color{red}{1}$$ for any $$n\in\mathbb{N}^+$$. We have just exploited the binomial theorem and the fact that $$\log(1+t)$$ is a holomorphic function in a neighbourhood of the origin such that $$g(0)=0$$. In particular it maps any simple closed contour around the origin in the region $$|t|<1$$ in a simple closed contour around the origin. This is the main idea of the Lagrange inversion theorem, see this brief outline.
$$\hspace{1cm}$$ This is how the contours $$|z|=\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}$$ are transformed via $$z\mapsto \log(z+1)$$.