Proof of binomial formula to extract coefficients of a generating function A generating function ends up rearranged into a form:
\begin{align}
(1+x+x^2+\dotsb)^n&=\frac{1}{(1-x)^n}\\[6px]
&= 1+\binom{1+n-1}{1}x+\binom{2+n-1}{2}x^2+\binom{3+n-1}{3}x^3\\
&\phantom{=\;1}+\dots+\binom{r+n-1}{r}x^r+\dotsb
\end{align}
used to extract coefficients.
I can't find a proof (some construction akin to Pascal's triangle, for example) of these coefficients working for all cases.
Probably, it is just a matter of knowing the term to include in the online search, and if this is the case, I'll be happy to delete the question.
 A: For $n=1$ it’s just a geometric series:
$$\frac1{1-x}=\sum_{n\ge 0}x^n\;.$$
Now suppose that
$$\frac1{(1-x)^n}=\sum_{k\ge 0}\binom{n-1+k}kx^k\;.$$
Then
$$\begin{align*}
\frac1{(1-x)^{n+1}}&=\frac1{(1-x)^n}\cdot\frac1{1-x}\\
&=\left(\sum_{k\ge 0}\binom{n-1+k}kx^k\right)\left(\sum_{k\ge 0}x^k\right)\\
&\overset{(1)}=\sum_{k\ge 0}\left(\sum_{j=0}^k\binom{n-1+j}j\right)x^k\\
&=\sum_{k\ge 0}\left(\sum_{j=0}^k\binom{n-1+j}{n-1}\right)x^k\\
&\overset{(2)}=\sum_{k\ge 0}\binom{n+k}nx^k\\
&=\sum_{k\ge 0}\binom{n+k}kx^k\;,
\end{align*}$$
as desired. Here $(1)$ is just a Cauchy product, and $(2)$ is a form of the hockey-stick identity.
A: The statement is obviously true for $n=1$; suppose it is for $n$; then we can see that
\begin{align}
(1+x+\dotsb)^{n+1}=\frac{1}{n}D\frac{1}{(1-x)^n}=
\frac{1}{n}\sum_{k\ge1}k\binom{k+n-1}{k}x^{k-1}=
\sum_{k\ge0}\frac{k+1}{n}\binom{k+n}{k+1}x^k
\end{align}
and it's just a matter of proving that
$$
\frac{k+1}{n}\binom{k+n}{k+1}=\binom{k+n}{k}
$$
Indeed,
$$
\frac{k+1}{n}\binom{k+n}{k+1}=
\frac{k+1}{n}\frac{(k+n)(k+n-1)\dotsm(k+n-(k+1)+1)}{(k+1)!}=
\binom{k+n}{k}
$$
A: Or,
if you write
$\dfrac1{(1-x)^n}
=(1-x)^{-n}
$,
you can use the 
generalized binomial series
$(1+x)^a
=\sum_{n=0}^{\infty} \binom{a}{n} x^n
$
where,
for any real (or complex) $a$,
$\binom{a}{n}
=\dfrac{\prod_{k=0}^{n-1}(a-k)}{n!}
$.
This converges whenever
$|x| < 1$.
Note that,
if $m$ is a positive integer,
$\begin{array}\\
\binom{-m}{n}
&=\dfrac{\prod_{k=0}^{n-1}(-m-k)}{n!}\\
&=(-1)^n\dfrac{\prod_{k=0}^{n-1}(m+k)}{n!}\\
&=(-1)^n\dfrac{\prod_{k=0}^{n-1}(m+(n-1-k))}{n!}\\
&=(-1)^n\dfrac{\prod_{k=0}^{n-1}(m+n-1-k)}{n!}\\
&=(-1)^n \binom{m+n-1}{n}\\
&=(-1)^n \binom{m+n-1}{m-1}\\
\text{so that}\\
(1-x)^{-m}
&=\sum_{n=0}^{\infty} \binom{-m}{n} (-x)^n\\
&=\sum_{n=0}^{\infty} (-1)^n \binom{m+n-1}{m-1} (-1)^n x^n\\
&=\sum_{n=0}^{\infty}  \binom{m+n-1}{m-1}  x^n\\
\end{array}
$
