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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

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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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