Gamma function integral How does the derivation of  this integral from the Gamma function is equivalent to (n-1)!?$$\Gamma(n)=\displaystyle\int_0^1\Bigg(\ln\frac1x\Bigg)^{n-1}dx$$ 

Using Gamma definition 
$$\Gamma(n)=\displaystyle\int_0^\infty e^{-u}u^{n-1}\ du$$
with substitution $e^{-u}=x$
$$\Gamma(n)=\displaystyle\int_0^1(\ln\frac1x)^{n-1}dx$$
