How to prove that generating function of series ($1 ,\binom{m}{1},\binom{m+1}{2},\binom{m+2}{3},...$ ) equals $\frac{1}{(1-z)^m}$ . How to prove that generating function of series ($1 ,\binom{m}{1},\binom{m+1}{2},\binom{m+2}{3},...$ ) equals $\frac{1}{(1-z)^m}$ . 
How do I do this? I know that I have to use some differentiation somewhere  but what series do I use to start?
 A: One way is to take $\frac{1}{1-z} = 1 + z + z^2 + \cdots$ and raise both sides to the $m$th power. To find the coefficient of $z^k$ in the resulting series, you need to solve a combinatorics problem (the stars and bars approach may help).
Another way is to start with $\frac{1}{1-z} = 1 + z + z^2 + \cdots$, differentiate both sides $m - 1$ times, and do some rearranging.
A: First a couple of preliminary notes before we get to the proof.

Lemma
$$\sum_{i=k}^n\binom{i}{k}=\binom{n+1}{k+1}$$

Proof of Lemma
The RHS counts the number of $k+1$ element subsets of $U=\{0\,\dotsc, n\}$. The RHS counts the same thing as the number of $k+1$ element subsets of $U$ with maximum element $i$ is $\binom{i}{k}$.$\blacksquare$
Note on Multiplication of Formal Power Series
In general if
$$
A(x)=\sum_{n\geq0}a_n x^n;\quad B(x)=\sum_{n\geq0}b_n x^n
$$
are two formal power series then their product is given by
$$
A(x)B(x)=\sum_{n\geq0}\left(\sum_{k=0}^na_kb_{n-k}\right)x^n.
$$
In particular since $(1-x)^{-1}=\sum_{n\geq 0} x^n$, it follows that
$$
A(x)(1-x)^{-1}=\sum_{n\geq0}\left(\sum_{k=0}^na_k\right)x^n\tag{1}
$$
The Problem

We claim that
  $$
(1-z)^{-m}=\sum_{n\geq0}\binom{m+n-1}{n}z^{n} \quad (m\ge 1)
$$

Proof by Induction
The base case holds since
$$
(1-z)^{-1}=\sum_{n\geq0} z^n=\sum_{n\geq0}\binom{1+n-1}{n}z^n
$$
as desired. Suppose that the result holds for all integers at most $m$. Then
$$
(1-z)^{-m-1}=(1-z)^{-m}\times(1-z)^{-1}=\sum_{n\geq0}\left(
\sum_{k=0}^n\binom{m+k-1}{k}
\right)
x^n
$$
by (1) and the induction hypothesis. But
$$
\sum_{k=0}^n\binom{m+k-1}{k}=\sum_{k=0}^n\binom{m+k-1}{m-1}=\sum_{i=m-1}^{n+m-1}\binom{i}{m-1}=\binom{n+m}{m}=\binom{m+n}{n}
$$
as desired where we have used the lemma.
A: Just use the fact that the coefficient of $z^n$ in the  power series expansion  near $0$ of a holomorphic function $f(z)$ is equal to $\;\dfrac{f^{(n)}(0)}{n!}$.
For $f(z)=\dfrac1{(1-z)^m}$, we see by an easy induction that
\begin{alignat}{2}
&&\biggl(\frac1{(1-z)^m}\biggr)^{\mkern-5mu(n)}&=\frac{m(m+1)\dotsm(m+n-1)}{(1-z)^{m+n}},\\
&\text{so }&f^{(n)}(0)&=m(m+1)\dotsm(m+n-1)=\frac{(m+n-1)!}{(m-1)!}\\
&\text{and eventually}\mkern-20mu &\frac{f^{(n)}(0)}{n!}&=\frac{(m+n-1)!}{(m-1)!\:n!}=\binom{m+n-1}{n}.
\end{alignat}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{k = 0}^{\infty}{m + k - 1 \choose k}z^{k} & =
\sum_{k = 0}^{\infty}\overbrace{{-\bracks{m + k - 1} + k - 1 \choose k}\pars{-1}^{k}}^{\ds{Negating\ \mbox{the Binomial}}}\,\,\,
z^{k} =
\sum_{k = 0}^{\infty}{-m \choose k}\pars{-z}^{k}
\\[5mm] & =
\bracks{\rule{0pt}{4mm}1 + \pars{-z}}^{\, -m} = \bbx{1 \over \pars{1 - z}^{m}}
\end{align}
