Residues of Gamma Function at Poles

Question

I am struggling to understand the result that the analytic continuation of the Gamma function, $\Gamma(z)$ has poles at $C \setminus \{0,-1,-2,.....\}$ and residues of $(-1)^k/k!$ at pole $k$. How may this be extended through the functional relation, to give the residuals of $\Gamma(z+1)$ or $z \Gamma(z)$ ?

Answer 1

Everything just follows from $\Gamma(z+1)=z\,\Gamma(z)$. For instance, since $\Gamma(1)=1$ and $\Gamma(z)=\frac{\Gamma(z+1)}{z}$, $z=0$ is a simple pole for the $\Gamma$ function with residue $1$. By induction, every negative integer is a simple pole for the $\Gamma$ function and

$$\text{Res}\left(\Gamma(z),z=-n\right) = \lim_{z\to -n}(z+n)\Gamma(z)=\lim_{z\to -n}\frac{\Gamma(z+n+1)}{z(z+1)\cdots(z+n-1)}=\color{red}{\frac{(-1)^n}{n!}}. $$ Similarly, for any negative integer $-n$ we have $$\text{Res}\left(\Gamma(z+1),z=-n\right)=\color{red}{\frac{(-1)^{n-1}}{(n-1)!}}.$$