Evaluating $\int_{0}^{1} \frac{\ln^{n} x}{(1-x)^{m}} \, \mathrm dx$ On another site, someone asked about proving that
$$ \int_{0}^{1} \frac{\ln^{n}x}{(1-x)^{m}} \, dx = (-1)^{n+m-1} \frac{n!}{(m-1)!} \sum_{j=1}^{m-1} (-1)^{j} s (m-1,j) \zeta(n+1-j), \tag{1} $$
where $n, m \in \mathbb{N}$, $n \ge m$, $m \ge 2$, and $s(m-1,j)$ are the Stirling numbers of the first kind.
My attempt at proving $(1)$:
$$ \begin{align}\int_{0}^{1} \frac{\ln^{n}x}{(1-x)^{m}} \, dx  &= \frac{1}{(m-1)!} \int_{0}^{1} \ln^{n} x \sum_{k=m-1}^{\infty} k(k-1)  \cdots (k-m+2) \  x^{k-m+1} \ dx \\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} k(k-1)  \cdots (k-m+2)  \int_{0}^{1} x^{k-m+1} \ln^{n} x \, dx \\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} k(k-1) \cdots (k-m+2) \frac{(-1)^{n} n!}{(k-m+2)^{n+1}}\\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} \sum_{j=0}^{m-1} s(m-1,j) \  k^{j}  \ \frac{(-1)^{n} n!}{(k-m+2)^{n+1}} \\ &= (-1)^{n}  \frac{n!}{(m-1)!} \sum_{j=0}^{m-1} s(m-1,j)  \sum_{k=m-1}^{\infty}   \frac{k^{j}}{(k-m+2)^{n+1}} \end{align} $$
But I don't quite see how the last line is equivalent to the right side of $(1)$. (Wolfram Alpha does say they are equivalent for at least a few particular values of $m$ and $n$.)

Here is another approach based on the beta function
$$ F = \int_{0}^{1}x^{\alpha}(1-x)^{-m} dx = \beta( \alpha+1,1-m ),\quad \alpha+1>0,\, 1-m>0. $$
Now, our integral $I$ can be evaluated as

$$ I = \lim_{\alpha \to 0} \frac{d^n}{d \alpha^n} \beta( \alpha+1,1-m ),\quad n>m-1,\,m\geq 2. $$

Here is a useful identity 

$$ 
{\partial \over \partial x} \mathrm{B}(x, y) = \mathrm{B}(x, y) \left( {\Gamma'(x) \over \Gamma(x)} - {\Gamma'(x + y) \over \Gamma(x + y)} \right) = \mathrm{B}(x, y) (\psi(x) - \psi(x + y)),  $$

where $\psi(x)$ is the digamma function. 
