analytic continuation of gamma function

The last of the parent entry (http://planetmath.org/GammaFunction) may be expressed as

$\mathrm{\Gamma }(z)={\displaystyle \frac{\mathrm{\Gamma }(z+n)}{z(z+1)(z+2)\mathrm{\cdots }(z+n-1)}}.$(1)

According to the standard definition

$$\mathrm{\Gamma }(z):={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{z-1}𝑑t,$$

the left hand side of (1) is defined only in the right half-plane  $\mathrm{\Re }z>0$, whereas the expression $\mathrm{\Gamma }(z+n)$ is defined and holomorphic for  $\mathrm{\Re }z>-n$ and thus the right hand side of (1) is holomorphic in the half-plane  $\mathrm{\Re }z>-n$ except the points

$$0,-1,-2,\mathrm{\dots },-(n-1)$$

where it has the poles of order 1.  Because the both sides of (1) are equal for $\mathrm{\Re }z>0$,  the left side of (1) is the analytic continuation of $\mathrm{\Gamma }(z)$ tothe half-plane  $\mathrm{\Re }z>-n$.  And since the positive integer $n$ can be chosenarbitrarily, the Euler’s $\mathrm{\Gamma }$-function has been defined analytically to the wholecomplex plane.

Accordingly, the gamma function is unambiguous and holomorphic everywhere in $ℂ$except in the points

$0,-1,-2,-3,\mathrm{\dots }$(2)

which are poles of order 1 of the function.  Hence, $\mathrm{\Gamma }(z)$is a meromorphic function.

For determining the residue of the function in the points (2), we rewrite the equation (1) as

$$\mathrm{\Gamma }(z)=\frac{\mathrm{\Gamma }(z+n+1)}{z(z+1)(z+2)\mathrm{\cdots }(z+n)}.$$

In the point  $z=-n$  we have

$$\mathrm{\Gamma }(z+n+1)=\mathrm{\Gamma }(1)=\mathrm{\hspace{0.33em}0}!=\mathrm{\hspace{0.33em}1},$$

which implies (see the rule in the entry coefficients of Laurent series) that

$$\mathrm{Res}(\mathrm{\Gamma };-n)=\frac{{(-1)}^n}{n!}.$$