Inverse Z-transform of $\frac{1}{(1-z^{-1})^2}$? 
What is the inverse Z-transform of $\frac{1}{(1-z^{-1})^2}$?

Title says it all. I have a one line solution but can't work out how to get there from tables or first principals.
Thanks!
 A: The Z-transform of a discrete time-domain signal $x = (x[n])$ is just the generating function $$X(z) = \sum \limits_{n=-\infty}^\infty x[n] z^{-n} .$$  Conversely, given a signal in the frequency-domain representation, i.e., given $X(z)$, the inverse Z-transform is simply its power series representation. 
In order to compute the inverse Z-transform of $X(z) = (1 - z^{-1})^{-2}$, we develop this as a power series in $z^{-1}$. This can be done by writing the Taylor series for the function $g(x) = (1-x)^{-2}$ and plugging in $x = z^{-1}$. Alternatively, we can also use the Newton binomial series formula: $$
(1-x)^{-(\beta+1)} =  \sum_{n=0}^{\infty} \binom{n+\beta}{\beta} x^n.
$$
Plugging in $x = z^{-1}$ and $\beta=1$, we get
$$
(1-x)^{-2} = \sum_{n=0}^{\infty} \binom{n+1}{1} z^{-n} = \sum_{n=0}^{\infty} (n+1) z^{-n}.
$$
Comparing with the definition of Z-transform, we conclude that the inverse Z-transform of $X(z)$ is  $$
x[n] = \begin{cases}
n+1, & n\geqslant 0, \\ 0, &n < 0.
\end{cases}
$$ 
Done!  $\qquad \diamond$

Using the properties. 
The above procedure is quite straightforward, but a bit tedious. Often it is possible to get the answer quicker by exploiting the properties of the Z-transform. To implement this approach effectively, one needs to be familiar with a table of Z-transforms of common series as well as a dictionary to translate elementary operations between the time and frequency domains. The Wikipedia article on Z-transforms contains such an extensive table of properties; my notation closely follows this page. 
Method 1. 
Given a signal $y[n]$ with Z-transform $Y(z)$, its accumulation $\sum \limits_{k=-\infty}^{n} y[k]$ has the Z-transform $\frac{Y(z)}{1 - z^{-1}}$. Now the delta function $\delta[n]$ (a unit spike at the origin and zero everywhere else) has the Z-transform $1$. Therefore, its accumulation given by
$$
u[n] = \begin{cases} 1, &n \geqslant 0, \\ 0, &n < 0, \end{cases}
$$
has the Z-transform $(1-z^{-1})^{-1}$. 
Applying the accumulation operation once again results in $(1-z^{-1})^{-2}$ in the frequency domain; in the time domain, we have
$$
\sum_{k=-\infty}^{n} u[k] = \sum_{k=-\infty}^n [k \geqslant 0] = \begin{cases} n+1, &n \geqslant 0, \\ 0, &n < 0, \end{cases}
$$
which is our desired inverse Z-transform.  $\qquad \diamond$
Alternate method using differentiation. 
We can replace the second accumulation operation with a differentiation operation. We again start from the signal
$$
u[n] = \begin{cases} 1, &n \geqslant 0, \\ 0, &n < 0,\end{cases}
$$
whose Z-transform is $U(z) = (1 - z^{-1})^{-1}$. Therefore the Z-transform of $n \cdot u[n]$ is given by
$$
- z \frac{d U(z)}{dz} = -z \cdot \frac{(-1)}{(1 - z^{-1})^{2}} \cdot \frac{1}{z^2} = \frac{z^{-1}}{(1-z^{-1})^2}. 
$$
Notice that we got what we wanted except for an extra $z^{-1}$. We can get rid of this by shifting the series by $1$ in the time domain: i.e., the Z-transform of $(n+1)\cdot u[n+1]$ is $(1-z^{-1})^{-2}$. Thus our inverse Z-transform is given by $(n+1) \cdot u[n+1]$. $\qquad \diamond$
A: This is the one that you find in the table:
$$
n\ \alpha^{n}\ u[n] <=> \frac{\alpha z^{-1}}{(1-\alpha z^{-1})^2}\ \ \ \ \ \ ROC:\ |z|>|\alpha|
$$
Then you find the shift property:
$$
x[n-k] <=> z^{-k}X(z)\ \ \ \ \ ROC:\ R_x
$$
What happens if you multiply RHS of transform by z?  then, LHS of transform has a shift.
$$
(n+1)\ \alpha^{n+1}\ u[n+1] <=> \frac{\alpha}{(1-\alpha z^{-1})^2}\ \ \ \ ROC: |z| > |\alpha| 
$$
when n=-1, (n+1) term equals zero. and u[n+1] = 1
when n=0,  (n+1) term equals one and u[n+1] = 1
now we notice that if we substitute u[n] in place of u[n+1], its the same because (n+1) is zero when n= -1.  Thus:
$$
(n+1)\ \alpha^{n+1}\ u[n] <=> \frac{\alpha}{(1-\alpha z^{-1})^2}\ \ \ \ ROC: |z| > |\alpha| 
$$
Then we divide the RHS of transform by alpha which corresponds to dividing by alpha on RHS of transform.
$$
(n+1)\ \alpha^{n}\ u[n] <=> \frac{1}{(1-\alpha z^{-1})^2}\ \ \ \ ROC: |z| > |\alpha|
$$
