Why does this relation of binomials hold? Why does $\sum_{m=n}^N m\binom {m-1}{n-1}=\sum_{m=n}^N n \binom mn=n \binom{N+1}{n+1}$ is true? Is there some special formula for it?
 A: The first equality is due to the equality in @OlivierOloa's answer.  The second one is a bit trick to obtain.


*

*Change $\binom nn$ to $\binom {n+1}{n+1}$.

*Group the first two term.

*Apply the equality $\binom nk + \binom n{k+1} = \binom {n+1}{k+1}$ to condense the leftmost two terms into one.

*Repeat (3) until only one term left.


\begin{align}
&\binom nn + \binom {n+1}n + \binom {n+2}n + \dots + \binom Nn \\
&= \left[\binom {n+1}{n+1} + \binom {n+1}n \right] + \binom {n+2}n + \dots + \binom Nn \\
&= \left[\binom {n+2}{n+1} + \binom {n+2}n\right] + \dots + \binom Nn \\
&= \cdots \\
&= \binom{N+1}{n+1}
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
\sum_{m = n}^{N}m{m - 1 \choose n - 1} & =
\sum_{m = n}^{N}m\bracks{z^{n - 1}}\pars{1 + z}^{m - 1} =
\bracks{z^{n - 1}}\sum_{m = n}^{N}m\pars{1 + z}^{m - 1}
\\[5mm] & =
\bracks{z^{n - 1}}\partiald{}{z}\sum_{m = n}^{N}\pars{1 + z}^{m} =
\bracks{z^{n - 1}}\partiald{}{z}\bracks{%
\pars{1 + z}^{n}\,{\pars{1 + z}^{N - n + 1} - 1 \over \pars{1 + z} - 1}}
\\[5mm] & =
\bracks{z^{n - 1}}\partiald{}{z}\bracks{%
\pars{1 + z}^{N + 1} - \pars{1 + z}^{n} \over z}
\\[5mm] & =
\bracks{z^{n - 1}}\partiald{}{z}\bracks{%
\sum_{k = 0}^{N + 1}{N + 1 \choose k}z^{k - 1} -
\sum_{k = 0}^{n}{n \choose k}z^{k - 1}}
\\[5mm] & =
\bracks{z^{n - 1}}\bracks{%
\sum_{k = 0}^{N + 1}{N + 1 \choose k}\pars{k - 1}z^{k - 2} -
\sum_{k = 0}^{n}{n \choose k}\pars{k - 1}z^{k - 2}}
\\[5mm] & =
\sum_{k = 0}^{N + 1}{N + 1 \choose k}\pars{k - 1}\delta_{n - 1,k - 2} -
\sum_{k = 0}^{n}{n \choose k}\pars{k - 1}\delta_{n - 1,k - 2}
\\[5mm] & =
{N + 1 \choose n + 1}n\bracks{0 \leq n + 1 \leq N + 1}\ -\
\underbrace{{n \choose n + 1}}_{\ds{=\ 0}}\
n\bracks{0 \leq n + 1 \leq N + 1}
\\[5mm] & = \bbx{n{N + 1 \choose n + 1}}
\end{align}
A: 
Is there some special formula for it?

One has
$$
m\binom {m-1}{n-1}=m\frac{(m-1)!}{(n-1)!(m-1-n+1)!}=n\frac{m!}{n!(m-n)!}=n \binom mn.
$$
A: Combinatorially:
$$m\binom {m-1}{n-1}=\underbrace{{m\choose 1}\binom {m-1}{n-1}}_{LHS}=\underbrace{\binom mn {n\choose 1}}_{RHS}=n\binom mn.$$
LHS: In a group of $m$ students, we select $1$ group representative. From the remaining $m-1$ students we select $n-1$ assistants. Note that in total $n$ students out of $m$ were selected.
RHS: In the same group of $m$ students, we select $n$ students ($1$ to be a group representative and the rest $n-1$ to be assistants). From these $n$ students we select $1$ group representative. 
