$$-\ln(1-x)=\sum_{n=1}^\infty \frac{x^n}{n}$$
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
When I work out the power series $\sum_{n=2}^\infty c_nx^n$ of $(\ln(1-x))^2$ using the Cauchy product of $\sum_{n=1}^\infty \frac{x^n}{n}$ and itself, I get $c_n=\sum_{k=1}^{n-1}\frac{1}{k}\frac{1}{n-k}$.
If I work out the Cauchy product of the power series of $-\ln(1-x)$ and $\frac{1}{1-x}$ and integrate it, I conclude that $c_n=\frac{2}{n}\sum_{k=1}^{n-1}\frac{1}{k}$.
I would like a more basic proof that $\sum_{k=1}^{n-1}\frac{1}{k}\frac{1}{n-k} = \frac{2}{n} \sum_{k=1}^{n-1}\frac{1}{k}$.
I failed at induction as well as trying to use the symmetry of the added terms.
A quite elementary way to show it is as follows:
First note that $$\sum_{k=1}^{n-1}\frac{1}{k}\frac{1}{n-k} = \frac{2}{n} \sum_{k=1}^{n-1}\frac{1}{k} \Leftrightarrow n\sum_{k=1}^{n-1}\frac{1}{k}\frac{1}{n-k} =2\sum_{k=1}^{n-1}\frac{1}{k} $$
Now, you may prove the equality on the right:
\begin{eqnarray*} n\sum_{k=1}^{n-1}\frac{1}{k(n-k)} & = & \sum_{k=1}^{n-1}\frac{n}{k(n-k)} \\ & = & \sum_{k=1}^{n-1}\frac{n-k+k}{k(n-k)} \\ & = & \sum_{k=1}^{n-1}\left( \frac{1}{k} + \frac{1}{n-k}\right) \\ & = & \sum_{k=1}^{n-1}\frac{1}{k} + \underbrace{\sum_{k=1}^{n-1}\frac{1}{n-k}}_{= \sum_{k=1}^{n-1}\frac{1}{k}} \\ & = & 2\sum_{k=1}^{n-1}\frac{1}{k} \end{eqnarray*}