# Gamma Difference Equation/Proof 2

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## Theorem

- $\map \Gamma {z + 1} = z \, \map \Gamma z$

## Proof

By Euler's form of the Gamma function:

\(\displaystyle \frac {\map \Gamma {z + 1} } {\map \Gamma z}\) | \(=\) | \(\displaystyle \paren {\frac 1 {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^{z + 1} } {1 + \frac {z + 1} n} } \div \paren {\frac 1 z \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^z} {1 + \frac z n} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac z {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \paren {\frac {\paren {1 + \frac 1 n} \paren {z + n} } {z + n + 1} }\) | after some more-or-less hairy algebra | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle z \lim_{m \mathop \to \infty} \frac {m + 1} {z + m + 1} = z\) |

$\blacksquare$