Find nth derivative of $\frac{x^{n}}{(1-x)^{2}}$, please? I need to find the nth derivative of $\frac{x^{n}}{(1-x)^{2}}$ for $0&lt;x&lt;1$

So far, I tried the same method used for $\frac{x^{n}}{1-x}$ and here's what I got:
\begin{equation}
\frac{x^{n}}{(1-x)^{2}}=x^{n}(1+2x+3x^{2}+4x^{3}+....)=(x^{n}+2x^{n+1}+3x^{n+2}+...)
\end{equation}
Take nth derivative to get:
\begin{align}
\frac{\partial^n }{\partial x^n} \frac{x^{n}}{(1-x)^{2}} \notag\\ = 
&amp; (n!+2(n+1)!x+3\frac{(n+2)!}{2!}x^{2}+...) \notag\\ =
&amp; n!(1+2(n+1)x+3\frac{(n+2)(n+1)}{2!}x^{2}+4\frac{(n+3)(n+2)(n+1)}{3!}x^{3}+..) 
\end{align}
Next would be finding a a function whose taylor expansion is the series in the parenthesis but I couldn't think of one. Any ideas for a function or for another method to do this? Thanks!
Write $x = 1-t$, so you're looking at $$(-1)^n \dfrac{d^n}{dt^n} \dfrac{(1-t)^n}{t^2} = \dfrac{d^n}{dt^n}  \sum_{j=0}^n {n \choose j} (-1)^{n+j}   t^{j-2}$$
The terms with $j \ge 2$ all vanish, and you're left with
$$ \dfrac{d^n}{dt^n} \left((-1)^n (t^{-2} - n t^{-1})\right)$$

From there it's easy.  You can write the answer as 
$$\dfrac{n!(nx+1)}{(1-x)^{n+2}}$$
You could write
$$\frac{\partial^n}{\partial x^n}x^n(1-x)^{-2}=\sum_{k=0}^n\binom{n}{k}\left(\frac{\partial^k}{\partial x^k}x^n\right)\left(\frac{\partial^{n-k}}{\partial x^{n-k}}(1-x)^{-2}\right)$$ and proceed from there. You'd have it as a finite sum.