If $k, m$ and $n$ are positive integers such that $n>m>k$, Prove that $$\binom{m}{0}\binom{n-m}{k}+\binom{m}{1}\binom{n-m}{k-1}+\dots+\binom{m}{k}\binom{n-m}{0}=\binom{n}{k}$$
I have re-written the statement using factorial notation and assume I must cancel out the $m$ to get to the RHS. However, I have not idea how to do this algebraically.
I also wrote it out in summation notation $$ \sum_{i = 0}^k \frac{m! }{i! (m-i)!} × \frac{(n-m)!}{(k-i)!(n -m-k + i)!} $$
but I don't know what to do with this information. Can someone guide me?
Combinatorial proof
The number $n \choose k$ tells us in how many ways we can choose $k$ elements out of $n$ elements. Now split those $n$ elements into two parts with $m$ and $n-m$ elements. To choose $k$ elements out of the $n$ elements we can choose $l \in \{0, \ldots, k\}$ elements out of the part with $m$ elements and $k-l$ elements out of the part with $n-m$ elements. For a given $l$ this results in ${m \choose l}{n-m \choose k-l}$ ways. Then summing over the different possible values of $l$ we get the result $$ \sum_{l=0}^{k} {m \choose l}{n-m \choose k-l} = {n \choose k}. $$
Algebraic proof
We split a binomial power into two factors with powers $m$ and $n-m$ respectively: $$ (x+1)^n = (x+1)^m (x+1)^{n-m} = \left( \sum_{l=0}^{m} {m \choose l} x^l \right) \left( \sum_{p=0}^{m} {n-m \choose p} x^p \right) \\= \sum_{l=0}^{m} \sum_{p=0}^{m} {m \choose l} {n-m \choose p} x^{l+p} %= \sum_{k=0}^{n} \left( \sum_{l=0}^{m} {m \choose l} {n-m \choose k-l} \right) x^{k} . $$ Now, $n \choose k$ is the coefficient for the $x^k$ term. This is found by only taking the terms where $p=k-l$: $$ {n \choose k} x^k = \sum_{l=0}^{m} {m \choose l} {n-m \choose k-l} x^{k}. $$