Power series, geometric series and differentiation 
By using the formula for the derivative of a power series, show that
  $$
\frac{1}{(1-z)^{k}}=\sum_{n=k-1}^{\infty}\left(\begin{array}{c}
{n} \\
{k-1}
\end{array}\right) z^{n-k+1}
$$
  for $|z|<1$ by using the expansion of the geometric series.

Things which I've done:
~Taken the derivative of the left hand side and expanded it using geometric series
~Expanded the left hand side using geometric series and then taken the derivative
None of the following seem to give me something comparable (that is, the expressions seemed too difficult to work with) to the right hand side. Any suggestions?
 A: Since $1/(1-z) = \sum_{n=0}^\infty z^n$
\begin{equation}
1/(1-z)^k = \left(\sum_{n=0}^\infty z^n\right)^k
= \sum_{n=0}^\infty a_n z^n
\end{equation}
where $a_n$ is the number of possible combinations of $(b_1,\ldots,b_k)$ such that
\begin{equation}
b_1 + \cdots + b_k = n
\end{equation}
where $b_i\geq0$.
It can be proven that
\begin{equation}
a_n = \left(\begin{array}{c}
n+k-1\\
k-1
\end{array}\right),
\end{equation}
hence
\begin{equation}
1/(1-z)^k 
= \sum_{n=0}^\infty
\left(\begin{array}{c}
n+k-1\\
k-1
\end{array}\right)
z^n
= \sum_{n=k-1}^\infty
\left(\begin{array}{c}
n\\
k-1
\end{array}\right)
z^{n-k+1},
\end{equation}
hence the proof!
A: Proof by induction, we have 
\begin{eqnarray*}
\frac{1}{(1-z)^{k}}=\sum_{n=k-1}^{\infty}\left(\begin{array}{c}
{n} \\
{k-1}
\end{array}\right) z^{n-k+1}.
\end{eqnarray*}
Differentiating gives
\begin{eqnarray*}
\frac{k}{(1-z)^{k+1}}=\sum_{n=k-1}^{\infty}\left(\begin{array}{c}
{n} \\
{k-1}
\end{array}\right) (n-k+1) z^{n-k}.
\end{eqnarray*}
So we just need to show that 
\begin{eqnarray*}
 \frac{(n-k+1) }{k}\left(\begin{array}{c}
{n} \\
{k-1}
\end{array}\right) = \binom{n}{k}
\end{eqnarray*}
which is easy.
