Gamma function integral How does the derivation of  this integral from the Gamma function is equivalent to (n-1)!?$$\Gamma(n)=\displaystyle\int_0^1\Bigg(\ln\frac1x\Bigg)^{n-1}dx$$ 
 A: Using Gamma definition 
$$\Gamma(n)=\displaystyle\int_0^\infty e^{-u}u^{n-1}\ du$$
with substitution $e^{-u}=x$
$$\Gamma(n)=\displaystyle\int_0^1(\ln\frac1x)^{n-1}dx$$
A: I assume you want the result only for positive integer $n$.  
Integrate by parts to get a reduction formula.
$$
\int_0^1\left(\log\frac{1}{x}\right)^{n-1} =
-\left.\frac{x}{n}\left(\log\frac{1}{x}\right)^n\;\right|_{x=0}^{x=1}
+\int_0^1\frac{1}{n}\left(\log\frac{1}{x}\right)^{n}
=\frac{1}{n}\int_0^1\left(\log\frac{1}{x}\right)^{n}
$$
And do the easy initial case $n=1$
$$
\int_0^1\left(\log\frac{1}{x}\right)^{0}dx = 1.
$$
So by induction,
$$
\int_0^1\left(\log\frac{1}{x}\right)^{n-1} = (n-1)!
$$ 
added
Computation for
$$
\left.\frac{x}{n}\left(\log\frac{1}{x}\right)^n\;\right|_{x=0}^{x=1} .
$$
For $x=1$, we have $\log(1) = 0$ so
$$
\frac{1}{n}\left(\log\frac{1}{1}\right)^n = 0.
$$
(Recall $n \ge 1$.)  
For $x=0$, we have
$$
\lim_{x \to 0^+}\frac{x}{n}\left(\log\frac{1}{x}\right)^n = 0
$$
since the $x$ goes to zero much, much faster than the $\log$ goes to $\infty$.
