Prove the following power series solution Prove the following power series equality:
$$\sum_{n\geq0} {n \choose k-1} x^{n} = x^{k-1}/(1-x)^k $$
I'm not sure if I am on the right track, but I tried answering this through induction on k. With base case  k = 1. I got  $$\sum_{n>=0}x^{n} = (1-x)^{-1} $$ which holds. When I try for m + 1 I get
$$ x^{m}/((1-x)^m(1-x)) $$ but im not sure how to continue from here. I was thinking of writing the bottom half of the fraction as a sum, but I'm not sure what I would do with the x^m term 
 A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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\begin{align}
\color{#f00}{\sum_{n \geq 0}{n \choose k - 1}x^{n}} & =
\sum_{n = k - 1}^{\infty}{n  \choose k - 1}x^{n} =
\sum_{n = 0}^{\infty}{n + k - 1 \choose k - 1}x^{n + k - 1} =
x^{k - 1}\sum_{n = 0}^{\infty}{n + k - 1 \choose n}x^{n}
\\[5mm] & =
x^{k - 1}\sum_{n = 0}^{\infty}{-n - k + 1 +n - 1 \choose n}\pars{-1}^{n}x^{n} =
x^{k - 1}\sum_{n = 0}^{\infty}{-k \choose n}\pars{-x}^{n}
\\[5mm] & =
x^{k - 1}\bracks{1 + \pars{-x}}^{-k} =
\color{#f00}{x^{k - 1} \over \pars{1 - x}^{k}}
\end{align}
