Prove $\sum_{i=0}^n\binom{m+i}{i}=\binom{m+n+1}{n}$ (another Hockey-Stick Identity?) Let $n$ be a nonnegative integer, and $m$ a positive integer. Could someone explain to me why the identity
$$
\sum_{i=0}^n\binom{m+i}{i}=\binom{m+n+1}{n}
$$
holds?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{i = 0}^{n}{m + i \choose i}} =
\sum_{i = 0}^{n}\pars{-1}^{i}{-m -i + i - 1 \choose i}
\\[5mm] = &\
\sum_{i = 0}^{n}\pars{-1}^{i}{-m - 1 \choose i} =
\sum_{i = 0}^{n}\pars{-1}^{i}\bracks{z^{i}}\pars{1 + z}^{-m - 1}
\\[5mm] = &\
\bracks{z^{0}}\pars{1 + z}^{-m - 1}
\sum_{i = 0}^{n}\pars{-\,{1 \over z}}^{i}
=
\bracks{z^{0}}\pars{1 + z}^{-m - 1}\,
{\pars{-1/z}^{n + 1} - 1 \over \pars{-1/z} - 1}
\\[5mm] = &\
\bracks{z^{0}}\pars{1 + z}^{-m - 1}\,
{\pars{-1}^{n + 1} - z^{n + 1} \over -1 - z}
\,{z \over z^{n + 1}}
\\[5mm] = &\
\bracks{z^{n}}\pars{1 + z}^{-m - 2}\,
\bracks{z^{n + 1} - \pars{-1}^{n + 1}} =
\pars{-1}^{n}\bracks{z^{n}}\pars{1 + z}^{-m - 2}
\\[5mm] = &\
\pars{-1}^{n}{-m - 2 \choose n} =
\pars{-1}^{n}\bracks{{m + 2 + n - 1\choose n}\pars{-1}^{n}}
\\[5mm] = &\ \bbx{m + n + 1 \choose n} \\ &
\end{align}
A: We have
$$\sum_{q=0}^n {m+q\choose q} = \sum_{q\ge 0} {m+q\choose q} [[0\le q\le n]]
\\ = \sum_{q\ge 0} {m+q\choose q} [z^n] \frac{z^q}{1-z}
= [z^n] \frac{1}{1-z}  \sum_{q\ge 0} {m+q\choose q} z^q
\\= [z^n] \frac{1}{1-z} \frac{1}{(1-z)^{m+1}}
= [z^n] \frac{1}{(1-z)^{m+2}}
= {n+m+1\choose n}.$$
