Prove Vandermond's identity: $$ \sum_{k=-\infty}^{+\infty} {r \choose m+k}{s \choose n-k} = {r+s \choose m+n}$$ given that: $\displaystyle {a \choose b} = 0$ if $b > a$ or $b < 0$
It's obvious that if $k < \max\{-m, n-s\}$ or $k > \min\{r-m, n\}$ then the whole term $\displaystyle {r \choose m+k}{s \choose n-k}$ is equal to zero. So, how do I make an argument that $k = [-m, n]$?
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = -\infty}^{\infty}{r \choose m + k}{s \choose n - k} & = \sum_{k = -\infty}^{\infty}{r \choose k}{s \choose m + n - k} = \sum_{k = -\infty}^{\infty}{r \choose k}\bracks{z^{m + n - k}}\pars{1 + z}^{s} \end{align}
where $\ds{\bracks{z^{m}}\mrm{f}\pars{z}}$ denotes the coefficient of $\ds{z^{m}}$ in an expansion of $\ds{\,\mrm{f}\pars{z}}$ in powers of $\ds{z}$. Obviously, $\ds{\bracks{z^{a - b}}\mrm{f}\pars{z} = \bracks{z^{a}}\braces{z^{b}\,\mrm{f}\pars{z}}}$.
Then, \begin{align} &\sum_{k = -\infty}^{\infty}{r \choose m + k}{s \choose n - k} = \sum_{k = -\infty}^{\infty}{r \choose k} \bracks{z^{m + n}}\braces{z^{k}\pars{1 + z}^{s}} \\[5mm] = &\ \bracks{z^{m + n}} \braces{\pars{1 + z}^{s}\sum_{k = -\infty}^{\infty}{r \choose k}z^{k}} = \bracks{z^{m + n}}\braces{\pars{1 + z}^{s}\pars{1 + z}^{r}} = \bracks{z^{m + n}}\pars{1 + z}^{r + s} \\[5mm] = &\ \bbx{\ds{r + s \choose m + n}} \end{align}
