Eulerian Integral of the Second Kind

$\displaystyle \Pi(z,n)$	$\textstyle \equiv$	$\displaystyle \int_0^n \left({1-{t\over n}}\right)^n t^{z-1}\,dt = n^z\int_0^1 (1-\tau)^n \tau^{z-1}\,d\tau$
	$\textstyle =$	$\displaystyle {n!\over z(z+1)\cdots (z+n)}n^z.$