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.

My usual naive plug in and assume
everything converges.
$\begin{array}\\
-n\int _0 ^1 x^{n-1} \log(1-x)dx
&=n\int _0 ^1 x^{n-1} \sum_{m=1}^{\infty}\dfrac{x^m}{m}dx\\
&=n\sum_{m=1}^{\infty}\dfrac{1}{m}\int _0 ^1 x^{n-1} x^mdx\\
&=n\sum_{m=1}^{\infty}\dfrac{1}{m}\int _0 ^1 x^{n+m-1}dx\\
&=n\sum_{m=1}^{\infty}\dfrac{1}{m}\dfrac1{n+m}\\
&=n\sum_{m=1}^{\infty}\dfrac{1}{m(n+m)}\\
&=n\sum_{m=1}^{\infty}\dfrac1{n}(\dfrac1{m}-\dfrac1{n+m})\\
&=\sum_{m=1}^{\infty}(\dfrac1{m}-\dfrac1{n+m})\\
&=\sum_{m=1}^{\infty}\dfrac1{m}-\sum_{m=1}^{\infty}\dfrac1{n+m}\\
&=\sum_{m=1}^{\infty}\dfrac1{m}-\sum_{m=n+1}^{\infty}\dfrac1{m}\\
&=\sum_{m=1}^{\infty}\dfrac1{m}-\sum_{m=n+1}^{\infty}\dfrac1{m}\\
&=\sum_{m=1}^{n}\dfrac1{m}\\
&=H_n\\
\end{array}
$
If you do the same thing
with $\log(1+x)$,
you get a similar result
but it cancels out
only for even $n$ -
I got 
$-n\int _0 ^1 x^{n-1} \log(1+x)dx\\
=\sum_{m=1}^{\infty}\dfrac{(-1)^{m+1}}{m}-(-1)^n\sum_{m=n+1}^{\infty}\dfrac{(-1)^{m+1}}{m}
$.
In any case,
$-n\int _0 ^1 x^{n-1} \log(1+x)dx
\to \ln(2)
$.
