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Factorials and the Gamma function

The above results give a simple proof of $ \Gamma(n+1) = n!$. Now

$\displaystyle \Gamma(n+1) = \int_0^\infty t^n {\rm e}^{-t} {\rm d}t
$

so, by (8),

$\displaystyle \Gamma(n+1) = \lim_{x \to \infty} {\rm e}^{-x} \sum_{r=0}^n (-1)^...
...)^{n-r+1}} - \frac{(-1)^n n!}{(-1)^{n+1}}
= -\frac{(-1)^n n!}{(-1)^{n+1}} = n!
$

since $ \lim_{x \to \infty} x^m {\rm e}^{-x} = 0$.

Alexander Frolkin 2001-06-02