Gamma Difference Equation

Theorem
The Gamma function satisfies

$$\Gamma(z+1)=z\Gamma(z) \ $$

for any $$z \ $$ which is not a nonpositive integer.

Proof
By Euler's form for the Gamma function,

$$\frac{\Gamma(z+1)}{\Gamma(z)}=\left({\frac{1}{z+1}\lim_{m\to\infty}\prod_{n=1}^m\frac{\left({1+\frac{1}{n}}\right)^{z+1}}{1+\frac{z+1}{n}}}\right)\div\left({\frac{1}{z}\lim_{m\to\infty}\prod_{n=1}^m \frac{\left({1+\frac{1}{n}}\right)^z}{1+\frac{z}{n}}}\right) \ $$

$$=\frac{z}{z+1}\lim_{m\to\infty}\prod_{n=1}^m\left({\frac{\left({1+\frac{1}{n}}\right)\left({z+n}\right)}{z+n+1}}\right) \ $$

$$=z\lim_{m\to\infty}\frac{m+1}{z+m+1}=z \ $$