Representing geometric series as sum of binomial coefficients I was learning generating functions, where faced following:

$$(1-x)^{-n}=\sum_{i=0}^\infty\binom{n+i-1}{i}x^i$$

I didn't get how is this arrived. I tried out to prove this to me. But failed.
Using binomial theorem, I can get that 
$$(1-x)^{-n}=\sum_{i=0}^\infty\binom{-n}{i}x^i$$
However I realize that, I am first time facing $-$ve $n$ in $\binom{-n}{i}$ and I dont know how to solve this. I tried but reached to something involving $-$ve factorial. Googling revealed that $(-n)!$ is somewhat undefined. Wolfram says $(-4)!$ is complex infinity. I also came across this thread which solves 
$$\binom{-4}{3}=\frac{(-4)_3}{3!}$$
I didnt get how this equality is arrived. 
Overall I am lost.
 A: There are several ways to derive the result; I’ll start with the one using the binomial theorem. What you have isn’t quite right: the $x$ term in $1-x$ is negative, so it should be
$$(1-x)^{-n}=\sum_{k\ge 0}\binom{-n}k(-x)^k=\sum_{k\ge 0}\binom{-n}k(-1)^kx^k\;.$$
For $\binom{-n}k$ you need to know the full definition of the binomial coefficient: for all real $x$ and non-negative integers $k$ we define
$$\binom{x}k=\frac{x^{\underline k}}{k!}=\frac{x(x-1)(x-2)\ldots(x-k+1)}{k!}\;;$$
you can check that this agrees with the more familiar definition when $x$ is a non-negative integer. Now we have
$$\begin{align*}
\binom{-n}k&=\frac{(-n)^{\underline k}}{k!}\\
&=\frac{(-n)(-n-1)(-n-2)\ldots(-n-k+1)}{k!}\\
&=(-1)^k\cdot\frac{n(n+1)(n+2)\ldots(n+k-1)}{k!}\\
&=(-1)^k\cdot\frac{(n+k-1)!}{k!(n-1)!}\\
&=(-1)^k\binom{n+k-1}k\;,
\end{align*}$$
so that
$$(1-x)^{-n}=\sum_{k\ge 0}(-1)^k\binom{n+k-1}k(-1)^kx^k=\sum_{k\ge 0}\binom{n+k-1}kx^k\;,$$
since $(-1)^k(-1)^k=(-1)^{2k}=1$.

Another way is to start from the geometric series
$$\frac1{1-x}=\sum_{k\ge 0}x^k\tag{1}$$
and differentiate repeatedly. If I differentiate $(1-x)^{-1}$ with respect $x$ repeatedly, I get $(1-x)^{-2}$, $2(1-x)^{-3}$, $6(1-x)^{-4}$, $24(1-x)^{-5}$, and in general I have
$$\frac{d^n}{dx^n}(1-x)^{-1}=\frac{n!}{(1-x)^{n+1}}\;;$$
this is easy to prove by induction on $n$.
Differentiating the righthand side of $(1)$ $n$ times with respect to $x$, I get
$$\sum_{k\ge 0}k(k-1)(k-2)\ldots(k-n+1)x^{n-k}\;.$$
Setting $\ell=k-n$, so that $k=\ell+n$, I can rewrite this as
$$\sum_{\ell\ge 0}(\ell+n)(\ell+n-1)\ldots(\ell+1)x^\ell=\sum_{\ell\ge 0}\frac{(\ell+n)!}{\ell!}x^\ell\;.$$
Putting the two pieces together, we see that
$$\frac{n!}{(1-x)^{n+1}}=\sum_{\ell\ge 0}\frac{(\ell+n)!}{\ell!}x^\ell\;,$$
or, after dividing by $n!$,
$$\frac1{(1-x)^{n+1}}=\sum_{\ell\ge 0}\binom{\ell+n}\ell x^\ell\;.$$
Now just replace $n$ by $n-1$ throughout to get
$$(1-x)^{-n}=\sum_{\ell\ge 0}\binom{\ell+n-1}\ell x^\ell\;,$$
as desired.
A: This is a comment
on Brian M. Scott's excellent answer.
I like to write
binomial coefficients
and similar products
directly in terms
of the product notation
and doing my manipulations
without any "..."s.
For example,
$\begin{array}\\
\text{Since}\\
\binom{n}{m}
&=\frac{\prod_{k=0}^{m-1}(n-k)}{m!}\\
\text{we have}\\
\binom{-n}{m}
&=\frac{\prod_{k=0}^{m-1}(-n-k)}{m!}\\
&=\frac{(-1)^m\prod_{k=0}^{m-1}(n+k)}{m!}\\
&=(-1)^m\frac{\prod_{k=0}^{m-1}(n+m-1-k)}{m!}
\qquad\text{(reversing the order of the product)}\\
&=(-1)^m\binom{n+m-1}{m}\\
\end{array}
$
