Prove that $\int_0^1 (x \log x)^{n-1}dx = (n-1)!/n^n$ by induction. I am trying to prove the following relation
$$ \int_0^1 (x \log x)^{n-1}\,\mathrm{d}x = (-1)^{n-1}\frac{(n-1)!}{n^{n}} \qquad \forall \ n \in \mathbb{N}\,,$$
by induction. It is clear that the formula holds for $n=1$. Assume that the expression holds for $n = k$. We want to show that this implies that the formula holds for $n = k + 1$. Then
$$ \int_0^{1} (x \log x)^{k} \,\mathrm{d}x = \, ? $$
However, I have trouble with the integration by parts, I can not find an $u$ and $v'$ to end up with $(x \log x)^{k}$ on the right-hand side. Any ideas or suggestions?
 A: $$\int_0^1 (x\log(x))^{n-1} dx = \left. \frac{1}{n} x^n \log(x)^{n-1} \right |_{x \to 0^+}^{x=1} - \frac{n-1}{n} \int_0^1 x^{n-1} \log(x)^{n-2} dx.$$ 
The boundary term simply vanishes. Going forward, you can repeat this procedure $n-1$ times; at that point, all the logs are gone, and then you will just need to integrate $x^{n-1}$ which is easy. The only issue is keeping track of the coefficients in front of the integral. (Incidentally, your desired result is not correct: if $n-1$ is odd then the result should be negative.)
This isn't really induction in the usual sense, though, in that the desired result is not used as an inductive hypothesis (because we don't actually remove the factor of $x$ along the way). Perhaps there is a way to do it in the more standard way. 
One idea is to differentiate $x\log(x)$ to obtain $\log(x)+1$, and then integrate $x^{n-1} \log(x)^{n-1}$ (switching notation so the desired thing has a power of $n$). This would mean:
$$\int x^n \log(x)^n dx = x\log(x) \int x^{n-1} \log(x)^{n-1} dx - \int dx (\log(x)+1) \int y^{n-1} \log(y)^{n-1} dy.$$
But when you proceed that way you have to deal with these undesirable iterated integrals, which get worse as you continue.
A: Actually there is a missing factor of $(-1)^{n-1}$ on the right hand side, which you can see from the fact that $\log x\lt0$ for $0\lt x\lt 1$.  
As for using induction to evaluate the integral, if we begin with the substition $x=e^{-u/n}$, we have
$$\int_0^1(x\log x)^{n-1}dx={(-1)^{n-1}\over n^n}\int_0^\infty u^{n-1}e^{-u}du$$
which leaves an easy integration-by-parts induction.
A: Let us prove a more general statement:
$$I(m,n):=\int_0^1 x^{m-1}\log^{n-1}(x)\,dx =(-1)^{n-1}\frac{(n-1)!}{m^{n}};\quad n \in \mathbb{N},\ m\ne0.\tag{1}$$
The original claim follows upon substitution $m=n$.
The equation (1) is evidently valid for $n=1$ and arbitrary $m\ne0$. Assume it is valid for $n>0$. It follows then that it is valid for $n+1$ as well:
$$ I(m,n+1)=\int_0^1 x^{m-1}\log^{n}(x)\,dx=\frac{1}{m}\int_0^1 \log^{n}(x) d x^{m}\\=
\underbrace{\left. \frac{1}{m} x^{m} \log^{n}(x) \right |_{0}^{1}}_{=0} - \frac{n}{m}\int_0^1 x^{m-1} \log^{n-1}(x) dx=- \frac{n}{m} I(m,n)\stackrel{I.H.}{=}(-1)^{n}\frac{n!}{m^{n+1}},$$
as claimed.
