Conjecture for a closed form of a parametric integral Let us assume that $n \in \mathbb{N}$. It is quite easy to prove that:
$$\int_{0}^{1} x^{n-1} \log (1-x) \, {\rm d}x = - \frac{\mathcal{H}_n}{n}$$
since if we use the series representation of $\mathcal{H}_m$ we have that:
\begin{equation} \mathcal{H}_m = \sum_{k=1}^{\infty} \left [ \frac{1}{k} - \frac{1}{m+k} \right ] \tag{1}\end{equation}
and it follows that our initial integral , call that $\mathcal{J}$ , 
\begin{align*} \mathcal{J} &=\int_{0}^{1} x^m \log(1-x) \, {\rm d}x \\   &= - \int_{0}^{1}x^m \sum_{n=1}^{\infty} \frac{x^n}{n} \, {\rm d}x \\   &= - \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1}x^{m+n} \, {\rm d}x\\   &= - \sum_{n=1}^{\infty} \frac{1}{n \left ( n+m+1 \right )}\\   &= - \sum_{n=1}^{\infty}\left[\frac{1}{(m+1)n} - \frac{1}{(m+1) \left ( m+n+1 \right )} \right]\\   &= - \frac{1}{m+1} \sum_{n=1}^{\infty} \left [ \frac{1}{n} - \frac{1}{n+m+1} \right ]\\  &\overset{(1)}{=} - \frac{\mathcal{H}_{m+1}}{m+1} \end{align*}
Now let us raise the exponent of $\log (1-x)$ by one. That is we now consider the integral
$$\int_{0}^{1} x^{n-1} \log^2 (1-x) \, {\rm d}x = \frac{1}{n} \left [ \mathcal{H}_n^2 + \mathcal{H}_n^{(2)} \right ]$$
The derivation is pretty much easy. What can we say for the integral 
$$\mathcal{J}(n, m) =\int_{0}^{1} x^{n-1} \log^m (1-x) \, {\rm d}x $$
I have a conjecture that states that it will involve a sum of $\mathcal{H}_n$ raised to the corresponding power of the log and a sum of $\mathcal{H}_n^{(m)}$. However, I am unable to make any progress for the general case. Diffing the Beta won't help us here. This is what I've done for the $m=2$ case. I just diffed the Beta twice. Any clever approach?
Also what we can say for the integral:
$$\mathcal{J}^*(n, m) =\int_{0}^{1} x^{n-1} \log^m (1+x) \, {\rm d}x $$
Well, one last thing one should note is that both integrals actually connect to Stirling Numbers of first kind. Maybe we could use the Taylor series expansion of $\log^m (1 \pm x)$ and reduce the problem to Stirling numbers. I am not that familiar though. 
 A: Not an answer, only some thoughts
Let $\displaystyle A_n^m = \int_0^{1} x^n \log^{m}(1-x) dx, \ A_n^0 = \frac{1}{n+1}$.
You get the following recurrence :
$$A_n^m =-\frac{m}{n+1}\sum_{k=0}^n A_k^{m-1}$$
but I admit I'm not sure what to do with it. There is the related but different recurrence ${n \choose m} = \sum_{k=0}^{n-1} {k \choose m-1}$ and the connexion $\binom{n}{m} = \sum_{k=0}^m \left[{ m \atop k} \right]\frac{n^k}{m!}$ to the the Stirling numbers, appearing in the Taylor expansion $\frac{\log^m (1+z)}{m!} = \sum_{k=m}^\infty (-1)^{k-m} \left[{k\atop m}\right] \frac{z^k}{k!} $ so that 
$\displaystyle A_n^m = \textstyle \int_0^1 x^n m!\sum_{k=m}^\infty (-1)^{k-m} \left[{k\atop m}\right] \frac{x^k}{k!}dx = \displaystyle \sum_{k=m}^\infty \frac{m! (-1)^{k-m}}{k! (n+k+1)} \left[{k\atop m}\right] $
Since $\sum_{m=0}^\infty z^m \frac{\log^m(1-x)}{m!} = e^{z \log(1-x)} = (1-x)^z$ we have the nice generating function
$\displaystyle F(z) = \sum_{m=0}^\infty \frac{A_n^m}{m!} z^m = \textstyle\int_0^1 x^n \sum_{m=0}^\infty z^m \frac{\log^m(1-x)}{m!} dx $ $= \int_0^1 x^n (1-x)^z dx = B(n+1,z+1)$
where $B(x,y)$ is the beta function

$\int_0^{1-\epsilon} \frac{m\log^{m-1}(1-x)}{x-1} dx = \log^m(1-\epsilon)$ and this is what we need for integrating by parts
$$A_n^m = \int_0^{1} x^n \log^{m}(1-x) dx$$ 
$\textstyle = \lim_{\epsilon \to 0} \frac{(1-\epsilon)^{n+1}}{n+1}\log^{m}(\epsilon)-\int_0^{1-\epsilon} \frac{x^{n+1}}{n+1} \frac{\log^{m-1}(1-x)}{x-1} dx $ $=  \lim_{\epsilon \to 0} \int_0^{1-\epsilon} \frac{m\log^{m-1}(1-x)}{(n+1)(x-1)} dx-\int_0^{1-\epsilon} \frac{x^{n+1}}{n+1} \frac{\log^{m-1}(1-x)}{x-1} dx$
$\textstyle = \frac{m}{n+1}\int_0^1 \frac{1-x^{n+1}}{x-1}\log^{m-1}(1-x)dx = -\sum_{k=0}^n \int_0^1 x^k\log^{m-1}(1-x)dx$
$$=-\sum_{k=0}^n A_k^{m-1}$$
