<s>[INST] Series expansion: $1/(1-x)^n$ What is the expansion for $(1-x)^{-n}$?
Could find only the expansion upto the power of $-3$. Is there some general formula?
 [/INST] 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:
$$
\begin{align}
\dfrac{1}{(1-x)} &= 1+x+x^2+x^3+x^4+x^5\dots+x^m+\dotsm\\
\dfrac{1}{(1-x)^2} &= 1+2x+3x^2+4x^3+5x^4\dots+mx^{m-1}+\dotsm\\ 
\dfrac{2\cdot 1}{(1-x)^3} &= 2+(3\cdot 2)x+(4\cdot 3)x^2+(5\cdot 4)x^3\dots+(m \cdot m-1)x^{m-2}+\dotsm\\ 
\vdots\\
\dfrac{(n-1)!}{(1-x)^n} &= \sum_{k=0}^\infty \dfrac{(k+n-1)!}{k!} x^k\\
\end{align}
$$
which can be simplified by dividing:
$$
\dfrac{1}{(1-x)^n} = \sum_{k=0}^\infty \dfrac{(k+n-1)!}{(n-1)!k!} x^k = \sum_{k=0}^\infty \binom{k+n-1}{n-1}x^k\\
$$
</s>