Does someone know, if the subsequent formula holds for $m \ge n \ge i \ge 1$ and if yes, can give a reference.
$$\sum_{k=i}^{m-n+i}\binom{k}{i}\binom{m-k}{n-i} = \binom{m+1}{n+1}$$
Thank you very much!
What you want is equation (5.26) on page 169 of
Concrete Mathematics (2nd edition)
by Ronald Graham, Donald Knuth, and Oren Patashnik.
Corrected: For integers $m,n\geq0$ and integers $\ell,q$ with $\ell+q\geq 0$
we have
$$\sum_{-q\leq k\leq \ell}{q+k\choose n}{\ell-k\choose m}={\ell+q+1\choose m+n+1}.$$
Let’s now substitute your variables $i\geq 0$ and $n-i\geq 0$ in the bottom
to obtain
$$\sum_{-q\leq k\leq \ell}{q+k\choose i}{\ell-k\choose n-i}={\ell+q+1\choose n+1}.$$
In fact, since you have assumed $i>0$, we get even more
$$\sum_{i-q\leq k\leq \ell}{q+k\choose i}{\ell-k\choose n-i}={\ell+q+1\choose n+1}.$$
That’s because ${q+k\choose i}=0$ when $q+k<i$.
Redefining the $k$ variable gives
$$\sum_{i\leq k\leq \ell+q}{k\choose i}{\ell+q-k\choose n-i}={\ell+q+1\choose n+1}.$$
Letting $m=\ell+q$ gives
$$\sum_{i\leq k\leq m}{k\choose i}{m-k\choose n-i}={m+1\choose n+1}.$$
Is this the same as your sum? Yes!
If $n=i$, then this is obvious.
When $n>i$, then ${m-k\choose n-i}=0$ for $k>m-n+i$ anyway.