Show that $\sum_{k=-\infty}^{+\infty} {r \choose m+k}{s \choose n-k} = {r+s \choose m+n}$. 
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 $k = [-m, n]$?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
