Prove the identity $\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$ 
Let $n,m \in \mathbb{N}$. Prove the identity $$\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$$

This seems very similar to Vandermonde identity, which states that for nonnegative integers we have $\sum^{m}_{k=0}\binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$. But, clearly this identity is somehow different from it. Any ideas?

For $t=0,1,\cdots, n$, $\binom{m+t}{m}$ is the coefficient of $x^m$ in the expansion of $(1+x)^{m+t}$.
Hence, 
$$\binom{m}{0}+\binom{m+1}{1}+\cdots +\binom{m+n}{n},$$
i.e.
$$\binom{m}{m}+\binom{m+1}{m}+\cdots +\binom{m+n}{m}$$
is the coefficient of $x^m$ in the expansion of
$$(1+x)^m+(1+x)^{m+1}+\cdots +(1+x)^{m+n}.$$
Here, we have
$$(1+x)^m+(1+x)^{m+1}+\cdots +(1+x)^{m+n}$$$$=\frac{(1+x)^m((1+x)^{n+1}-1)}{x}=\frac{(1+x)^{m+n+1}}{x}-\frac{(1+x)^m}{x}.$$
Since there is no term of $x^m$ in $\frac{(1+x)^m}{x}$, what we want is the coefficient of $x^{m+1}$ in $(1+x)^{m+n+1}$, i.e. $\binom{m+n+1}{m+1}=\binom{m+n+1}{n}$.
