TITLE: Show that $\int_0^1 (\ln x)^n dx =(-1)^n n!$
QUESTION [5 upvotes]: How can I prove the following: 
$$\int_0^1 (\ln x)^n dx =(-1)^n n!$$
where $n$ is an integer and $n>0$?
By using partial integration, I started by finding a reduction formula
$$
\begin{align*}
I_n &= \int (\ln x)^n dx \\ &= x(\ln x)^n - nI_{n-1}
\end{align*}
$$
however the bounds 0 and 1 complicate things seeing as $\ln x \to -\infty \quad \mathrm{as} \quad x \to 0^+$

REPLY [4 votes]: You can use the $\; \Gamma\;$function to evaluate the integral.
Substituting $x=e^t \;$
$$\  \int_0^1 {(lnx)^ndx}=\int_{- \infty}^0 {t^n}e^tdt$$
Substituting $\, p=-t \,$; $$ \int_{- \infty}^0 {t^n}e^tdt=(-1)^n\int_{0}^{\infty}{t^ne^{-p}dp}=(-1)^n n!$$