proving combinatorics identity - $\sum_{k=0}^m{n-k \choose m-k}={n+1 \choose m}$ 
Prove that for every $n \ge m \ge 1 , \sum_{k=0}^m{n-k \choose m-k}={n+1 \choose m}$ 

I've tried saying that the RHS represents the number of binary series with m "1" 's and n+1-m "0"'s, but I couldn't find out what k represents at the LHS.
Thanks

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\color{#f00}{\sum_{k = 0}^{m}{n - k \choose m - k}} & =
\sum_{k = 0}^{m}{-n + k + m - k - 1 \choose m - k}\pars{-1}^{m - k} =
\pars{-1}^{m}\sum_{k = 0}^{m}\pars{-1}^{k}{m - n - 1 \choose m - k}
\\[3mm] & =
\pars{-1}^{m}\sum_{k = 0}^{m}\pars{-1}^{k}\ \overbrace{%
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{m - n - 1} \over z^{m - k + 1}}\,
{\dd z \over 2\pi\ic}}^{\ds{=\ {m - n - 1 \choose m - k}}}
\\[3mm] & =
\pars{-1}^{m}\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{m - n - 1} \over z^{m + 1}}
\sum_{k = 0}^{m}\pars{-z}^{k}\,{\dd z \over 2\pi\ic}
\\[3mm] & =
\pars{-1}^{m}\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{m - n - 1} \over z^{m + 1}}
{\pars{-z}^{m + 1} - 1\over \pars{-z} - 1}\,{\dd z \over 2\pi\ic}
\\[3mm] & =
\overbrace{%
\oint_{\verts{z} = 1^{-}}\pars{1 + z}^{m - n - 2}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ 0}}\ +\
\pars{-1}^{m}\ \overbrace{%
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{m - n - 2} \over z^{m + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{=\ {m - n - 2 \choose m}}}
\\[3mm] & =
\pars{-1}^{m}{m - n - 2 \choose m} = \pars{-1}^{m}{-m + n + 2 + m - 1 \choose m}
\pars{-1}^{m} =
\color{#f00}{{n + 1 \choose m}}
\end{align}
