How to prove that $ -n \int _0 ^1 x^{n-1} \log(1-x)dx$ equals the $n$-th harmonic number? From (Almost) Impossible Integrals, Sums, and Series section 1.3:

$$H_n = -n\int _0 ^1 x^{n-1} \log(1-x)dx$$

The proof of which was appetizingly difficult. I was unable to answer the follow-up challenge question, and do not have access to the given solution. It asks:

Is it possible to [prove this equality] with high school knowledge only (supposing we know and use the notation of the harmonic numbers)?

I've been fiddling with the integral on the right for an hour, but that $\log$ really throws a wrench in my plans.

If you call the RHS $I_n$, then
\begin{align}I_n-I_{n-1}&=\int_0^1((n-1)x^{n-2}-nx^{n-1})\log(1-x)\,dx\\
&=\left[(x^{n-1}-x^n)\log(1-x)\right]_{x=0}^1
+\int_0^1\frac{x^{n-1}-x^n}{1-x}\,dx
\end{align}
on integration by parts.
Then
$$\lim_{x\to1}(x^{n-1}-x^n)\log(1-x)=\lim_{x\to1}(1-x)\log(1-x)=
\lim_{y\to0}y\log y=-\lim_{t\to\infty}te^{-t}=0$$
and the integral reduces to $\int_0^1x^{n-1}\,dx=1/n$. Therefore
$I_n-I_{n-1}=1/n$. Similarly $I_1=1$, using integration by parts.
I'd count all of this as A-level level maths.
