How to prove this complex binomial power series identity? I am trying to prove the following:
For $k \in \mathbb N$ and complex $z$ such that $|z|<1$:
$$ {1 \over (1-z)^{k +1}} = \sum_{n \ge 0} {n+k \choose k} z^n$$
But I can't do it. My first idea was to try induction over $k$ and I can do the base case $k=0$. But the induction step is difficult. 

Is there an easier way to prove this or do I have to do it by
  induction?

 A: Induction is fairly straightforward. Let $P_k$ be the statement $$\dfrac1{(1-z)^{k+1}} = \sum_{n \geq 0} \dbinom{n+k}k z^n$$
Consider the base case.
Recall that the geometric series
$$\dfrac1{1-z} = \sum_{n \geq 0}z^n$$
converges for $\vert z \vert < 1$. This validates the inductive step for $k=0$. Assume that induction is true for $k=m$, i.e., we have
$$\dfrac1{(1-z)^{m+1}} = \sum_{n \geq 0} \dbinom{n+m}m z^n$$
Differentiate both sides. This gives us
$$\dfrac{(m+1)}{(1-z)^{m+2}} = \sum_{n \geq 0} \dbinom{n+m}m n \cdot z^{n-1}$$
This gives us
\begin{align}
\dfrac1{(1-z)^{m+2}} & = \sum_{n \geq 0} \dbinom{n+m}m \dfrac{n}{m+1} \cdot z^{n-1} = \sum_{n \geq 1} \dfrac{(n+m)!}{n!m!} \dfrac{n}{m+1} \cdot z^{n-1}\\
& = \sum_{n \geq 1} \dfrac{(n+m)!}{(n-1)!(m+1)!} z^{n-1} = \sum_{n \geq 0} \dfrac{(n+m+1)!}{n!(m+1)!} z^{n} = \sum_{n \geq 0} \dbinom{n+m+1}{m+1}z^n
\end{align}
Hence, assume $P_m$, we have $P_{m+1}$ to be true.
A: You can this via Taylor's theorem with the integral form of the remainder; what this actually amounts to is a lot of integration by parts:
$$ \frac{1}{(1-z)^{k+1}} - \sum_{n=0}^{m} \binom{n+k}{n} = \int_0^z \frac{(z-t)^{m}}{m!} \frac{1}{(1-t)^{(k+1)+(m+1)}} \frac{(k+m+1)!}{k!} \, dt, $$
which can be shown by induction on $m$. Of course, the factorials on the right-hand side amount to
$$ (k+m+1)\frac{(k+m)!}{k!m!} = (k+m+1)\binom{m+k}{m}. $$
A: There is another approach using complex variables that you may want to
study for future  reference. Admittedly what follows is  an example of
an extreme proof of a simple statement.
Suppose we are trying to evaluate
$$\sum_{n\ge 0} {n+k\choose k} z^n$$
where $|z|<1.$

Put
$${n+k\choose k}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{n+k}}{w^{k+1}} \; dw.$$
This gives for the sum
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w^{k+1}} 
\sum_{n\ge 0} z^n (1+w)^n\; dw.$$
The geometric series converges for $|z(1+w)|<1$ or
$|1+w|<1/|z|$ to give
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w^{k+1}} 
\frac{1}{1-(1+w)z} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w^{k+1}} 
\frac{1}{1-z-zw} \; dw.$$
On the other hand we have
$$\frac{1}{(1-z)^{k}} =
\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{1}{w^{k+1}}\frac{1}{1-w/(1-z)}\; dw.$$
Now put $1/w = (1+u)/u$ so that $w=u/(1+u)=1-1/(1+u)$ and
$dw = 1/(1+u)^2 \; du$ to obtain
$$\frac{1}{2\pi i}
\int_{|u|=\epsilon} 
\frac{(1+u)^{k+1}}{u^{k+1}}\frac{1}{1-u/(1+u)/(1-z)}
\frac{1}{(1+u)^2}\; du
\\ = \frac{1}{2\pi i}
\int_{|u|=\epsilon} 
\frac{(1+u)^{k+1}}{u^{k+1}}
\frac{(1+u)(1-z)}{(1+u)(1-z)-u} \frac{1}{(1+u)^2} \; du
\\ = \frac{1}{2\pi i}
\int_{|u|=\epsilon} 
\frac{(1+u)^{k}}{u^{k+1}}
\frac{1-z}{1-z-uz} \; du.$$
This shows that
$$\frac{1}{(1-z)^{k}} =
(1-z)\sum_{n\ge 0} {n+k\choose k} z^n$$
or
$$\frac{1}{(1-z)^{k+1}} = \sum_{n\ge 0} {n+k\choose k} z^n,$$
QED.
