The identity with sum of binomial coefficients Is this formula true?
$$
\sum_{m=k}^n {m \choose k}={n+1 \choose k+1}
$$
If yes how to prove it?
 A: Combinatorial proof
Let $A=\{0,\cdots,n\}$.  This set has $n+1$ elements, so $\begin{pmatrix}n+1\\k+1\end{pmatrix}$ counts the number of subsets of size $k+1$.
On the other hand, for any subset of size $k+1$, let $m$ be its largest element.  It is certainly true that $k\leq m\leq n$.  Since $\begin{pmatrix}m\\k\end{pmatrix}$ is the number of subsets of $\{0,\cdots,m-1\}$ with $k$ elements, it is also the number of subsets of $\{0,\cdots,n\}$ with $k+1$ elements and $m$ as the largest element.
These two counts count the same thing, and, therefore, are equal.
A: there is also a straightforward inductive proof based on
$$
\sum_{m=k}^{n+1} {m \choose k}= \sum_{m=k}^n {m \choose k}+{n+1 \choose k} = {n+1 \choose k+1}+{n+1 \choose k}={n+2 \choose k+1}
$$
A: Observe that when we introduce the integral representation
$${m\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{k+1}}\; dz$$
we get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{m=k}^n
\frac{(1+z)^m}{z^{k+1}}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}}
\sum_{m=k}^n
(1+z)^m \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}}
\frac{(1+z)^{n+1}-(1+z)^k}{1+z-1} \;dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+2}}
\left((1+z)^{n+1}-(1+z)^k\right) \; dz.$$
This has two pieces, the first of which is
$${n+1\choose k+1}$$
and the second, which is zero,
thus concluding the proof.
This MSE link has another computation in the same spirit.
