<s>[INST] 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?
 [/INST] 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.
</s>