Factorials and the Gamma function

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

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

so, by (8),

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

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