Alpha, beta, gamma

A more pedestrian solution could be:

Let: $a_n = \int_0^\infty x^{n} e^{-x}\,dx$

By integration by parts: $a_n = [-x^{n} e^{-x}]_0^{\infty } - \int_0^\infty -nx^{n-1} e^{-x}\,dx$ So that for n >0: $a_n = na_{n-1}$ since $a_0 = 1$ : $a_n = n!$

This is the gamma function:

$\Gamma(t) = \int_0^\infty x^{t-1} e^{-x}\,dx$

where t = 6.

One nice property of the gamma function is that for an integer, $n$ :

$\Gamma(n) = (n-1)!$

So, $\Gamma(6) = (6-1)! = \boxed{120}$