idk if its (n-1)x(n-2)! ....
or (n-1)x(n)! .....
Videos
by plugging in 
into the expression for 
. 
so 
I realize this is an old thread, but I wanted to expand on the above answers on how to derive the formula for anyone else that might come along. Starting with the geometric series and taking successive derivatives:
$$ \begin{align} \dfrac{1}{(1-x)} &= 1+x+x^2+x^3+x^4+x^5\dots+x^m+\dotsm\\ \dfrac{1}{(1-x)^2} &= 1+2x+3x^2+4x^3+5x^4\dots+mx^{m-1}+\dotsm\\ \dfrac{2\cdot 1}{(1-x)^3} &= 2+(3\cdot 2)x+(4\cdot 3)x^2+(5\cdot 4)x^3\dots+\big(m \cdot (m-1)\big)x^{m-2}+\dotsm\\ \vdots\\ \dfrac{(n-1)!}{(1-x)^n} &= \sum_{k=0}^\infty \dfrac{(k+n-1)!}{k!} x^k\\ \end{align} $$
which can be simplified by dividing:
$$ \dfrac{1}{(1-x)^n} = \sum_{k=0}^\infty \dfrac{(k+n-1)!}{(n-1)!k!} x^k = \sum_{k=0}^\infty \binom{k+n-1}{n-1}x^k\\ $$
Yes its the binomial expansion for any index.
$(1-x)^{-n} = (-x)^{0} + -n(-x)^{1}+ \dfrac{-n(-n-1)}{2!}(-x)^{2} + ...$
which simplifies to ..
$(1-x)^{-n} = 1 + nx+ \dfrac{n(n+1)}{2!}(x)^{2} + \dfrac{n(n+1)(n+2)}{3!}(x)^{3} ...$
ie,
$(1-x)^{-n} = 1 + nx+ {n+1\choose 2}(x)^{2} + {n+2\choose 3}(x)^{3} ...$
Binomial expansion for any index is generalization of binomial theorem for positive integral index:
$$(1+x)^n = {n\choose 0} + {n\choose 1}x + {n\choose 2}x^2 + ...$$
Well, as I understand it, we could write the binomial expansion as:
$$(1-x)^n= \sum_{k=0}^{n} \binom n k 1^{n-k}\,(-x)^k$$ $$\binom{n}{0}1^n (-x)^0 + \binom n 1 1^{n-1} (-x)+ \binom n 2 1^{n-2}(-x)^2 + \binom n 3 1^{n-3}(-x)^3 \ldots$$
which simplifies to $$1-nx+\frac{n(n-1)}{2!}\cdot x^2 -\frac{n(n-1)(n-2)}{3!} \cdot x^3 \ldots$$
Which is the answer everyone else has given.
It is
$= 1+ n(-x) + (n(n-1)(-x)^2) / 2! + (n(n-1)(n-2)(-x)^3) / 3!$
When you expand a factorial say 3(n+1)
Is this the correct way to do it
(3n+1)(3n+2)(3n+3)
Or
(3n+1)(n+2)(n+3)
Or am I getting it completely wrong thank you.