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

Question

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.

Answer 1

$$\operatorname*{Res}_{z=0}\frac{1}{(1-e^{-z})^n}=\frac{1}{2\pi i}\oint\frac{dz}{(1-e^{-z})^n}\stackrel{z\mapsto \log(1+t)}{=}\frac{1}{2\pi i}\oint\frac{dt}{(1+t)\left(\frac{t}{1+t}\right)^n} $$ 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.

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This is how the contours $|z|=\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}$ are transformed via $z\mapsto \log(z+1)$.