Poles of Gamma Function

Theorem
The gamma function $$\Gamma :\C \to \C \ $$ is analytic throughout the complex plane except at $$\left\{{0,-1,-2,-3, \dots }\right\} \ $$ where it has simple poles.

Proof
We examine the equivalent form of the Gamma function given by Weierstrass:

$$\frac{1}{\Gamma(z)} = ze^{\gamma z} \prod_{n=1}^\infty \left({ \left({ 1+\frac{z}{n} }\right) e^{\frac{-z}{n}} }\right) \ $$

The terms of the product clearly do not tend to zero, and so the product is only zero if one of the terms is zero; this occurs when $$1+\frac{z}{n} = 0 \ $$, which occurs at $$z \in \left\{{-1,-2, \dots }\right\} \ $$. We also have the expression outside the product to consider; since the exponential function is never 0, this expression is zero whenever $$z=0 \ $$. Hence $$\frac{1}{\Gamma(z)} = 0 \ $$ when, and only when, $$z \in \left\{{0,-1,-2,-3, \dots }\right\} \ $$.