Differential calculus prove of $\sum_{i=0}^n {{i}\choose{ k}} = {{n+1}\choose{ k+1}}$ I asked in previous question for combinatorial proof of $\sum_{i=0}^n {{i}\choose{ k}} = {{n+1}\choose{ k+1}}$ but I heard it's possible to prove it using differential calculus (for sums). How to do that?
 A: $$
\begin{align}
\sum_{k=0}^n\sum_{j=0}^n\binom{j}{k}x^k
&=\sum_{j=0}^n(1+x)^j\tag{1}\\
&=\frac{(1+x)^{n+1}-1}{x}\tag{2}\\[3pt]
&=\sum_{k=0}^n\binom{n+1}{k+1}x^{k}\tag{3}
\end{align}
$$
Explanation:
$(1)$: Binomial Theorem
$(2)$: Sum of a Geometric Series
$(3)$: Binomial Theorem
Equate the powers of $x$.
A: Let $\mathbb{D}$ be the open unit disc $\big\{z\in\mathbb{C}\,\big|\,|z|<1\big\}$ in $\mathbb{C}$.  Note that $(1-z)^{-k-1}=\displaystyle\sum_{r=0}^\infty\,\binom{r+k}{k}\,z^r$ for $z\in\mathbb{D}$.  Consider $f(z):=\displaystyle\sum_{j=0}^{n-k}\,\frac{(1-z)^{-k-1}}{z^{j+1}}$ for all $z\in \mathbb{D}\setminus\{0\}$.  
Let $\gamma$ be a closed smooth curve in $\mathbb{D}$ that encircles the origin once in the counterclockwise direction.  Then, $$\frac{1}{2\pi\text{i}}\,\oint_\gamma\,f(z)\,\text{d}z=\sum_{j=0}^{n-k}\,\binom{j+k}{k}=\sum_{i=k}^n\,\binom{i}{k}\,,$$
as $\displaystyle\frac{1}{2\pi\text{i}}\,\oint_\gamma\,\frac{(1-z)^{-k-1}}{z^{j+1}}\,\text{d}z=\binom{j+k}{k}$ for all $j=0,1,2,\ldots,n$.
On the other hand, for any $z\in\mathbb{D}\setminus\{0\}$, we have
$$\begin{align}
f(z)&=(1-z)^{-k-1}\,\sum_{j=0}^{n-k}\,\frac{1}{z^{j+1}}=\frac{(1-z)^{-k-1}}{z}\left(\frac{1-\frac{1}{z^{n-k+1}}}{1-\frac{1}{z}}\right)
\\
&=\frac{(1-z)^{-k-2}\left(1-z^{n-k+1}\right)}{z^{n-k+1}}=\frac{(1-z)^{-k-2}}{z^{n-k+1}}-(1-z)^{-k-2}\,.
\end{align}$$
Hence, $\displaystyle\frac{1}{2\pi\text{i}}\,\oint_{\gamma}\,f(z)\,\text{d}z=\frac{1}{2\pi\text{i}}\,\oint_{\gamma}\,\frac{(1-z)^{-k-2}}{z^{n-k+1}}\,\text{d}z$ because $\displaystyle\oint_\gamma\,(1-z)^{-k-2}\,\text{d}z=0$.  As $$(1-z)^{-k-2}=\displaystyle\sum_{r=0}^\infty\,\binom{r+k+1}{k+1}\,z^r$$ for all $z\in\mathbb{D}$, we obtain
$$\frac{1}{2\pi\text{i}}\,\oint_\gamma\,f(z)\,\text{d}z=\binom{(n-k)+k+1}{k+1}=\binom{n+1}{k+1}\,.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\sum_{i = 0}^{n}{i \choose k} & =
\sum_{i = 0}^{n}\
\overbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{i} \over z^{k + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{=\ {i \choose k}}}\ =\
\oint_{\verts{z} = 1}{1 \over z^{k + 1}}\sum_{i = 0}^{n}\pars{1 + z}^{i}
\,{\dd z \over 2\pi\ic}
\\[3mm] & =
\oint_{\verts{z} = 1}{1 \over z^{k + 1}}
{\pars{1 + z}^{n + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic}\ =\
\overbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{n + 1} \over z^{k + 2}}
\,{\dd z \over 2\pi\ic}}^{\ds{=\ {n + 1 \choose k + 1}}}\ -\
\overbrace{\oint_{\verts{z} = 1}{1 \over z^{k + 2}}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ \delta_{k,-1}}}
\\[3mm] & = \fbox{$\ds{{n + 1 \choose k + 1} - \delta_{k,-1}}$}
\end{align}
