inverse z-transform of $\frac{1}{(1-z^{-1})^{2}}$

Question

how to find inverse-z-transform of:

$$ x[n] = Z-Transform\left\{ \frac{1}{(1-z^{-1})^{2}} \right\} $$

I could convert it to convolution... but then i'm stuck with solving convolution...which i'm not really sure how to perform either.

A few z-transform table definitions I'm looking at to solve the problem:

$$ \begin{aligned} X(z) &= \sum_{n=-\infty}^{\infty} x[n] z^{-n} \\ \\ \alpha^{n}\ u[n] &<=> \frac{1}{1-\alpha z^{-1}} \\ \\ x[n]*y[n] &<=> X(z)\ Y(z) \\ \\ x[n-k] &<=> z^{-k}X(z) \end{aligned} $$

Answer 1

for positive m: $$ \left(\begin{array}{c} n + m - 1 \\ m - 1 \end{array} \right)\ a^n\ u[n] <=> \frac{1}{(1-a z^{-1})^m} $$

$$ \left(\begin{matrix}n\\k\\\end{matrix}\right)=\frac{n\left(n-1\right)\ldots\left(n-k+1\right)}{k\left(k-1\right)\ldots1} $$

for m=2:

$$ (n+1)\ a^{n}\ u[n] <=> \frac{1}{(1-az^{-1})^2}\ \ \ \ ROC: |z| > |a| $$