We have the Riemann zeta function
$$ \zeta(s) = \frac{\pi}{2} \int_{\gamma} \frac{dz}{2\pi i} \, \frac{1}{z^s} \cot(\pi z) $$
and I want to use the Residue theorem in order to understand why it is true that
$$ \zeta(s) = -\frac{\pi}{2}\frac{1}{s!} \frac{d^{(s)}(z\cot(\pi z)) }{dz^s} \Big|_{z=0} $$
I understand the pole is at $z=0$ and that $\cot$ has no poles. If this is a "simple" pole at $z=0$ which formula am I supposed to use to find the residues?
We assume that $s$ is a positive integer. Recall that the residue of a function $f$ at $z=a$ for a pole of order $n+1$ can be expressed as
$$\text{Res}\left(f(z),z=a\right)=\frac{1}{n!}\lim_{z\to a}\frac{d^n}{dz^n}\left((z-a)^{n+1}f(z)\right)$$
Now, note that at $z=0$, $f(z)=\frac{\cot(\pi z)}{z^s}$ has a pole of order $s+1$. Therefore, the residue $f$ is given by
$$\text{Res}\left(\frac{\cot(\pi z)}{z^s},z=0\right)=\frac{1}{s!}\lim_{z\to 0}\frac{d^s}{dz^s}\left(z^{s+1}\frac{\cot(\pi z)}{z^s}\right)$$
Therefore, if $\gamma$ is a contour over which $|z|<1$, then
$$\oint_\gamma \frac{\cot(\pi z)}{z^s}\,dz=2\pi i \frac{1}{s!}\lim_{z\to 0}\frac{d^s}{dz^s}\left(z\cot(\pi z)\right)$$