Proof that $ \sum_k (-1)^k \binom{n-k}{n-m} \binom{m}{k} = \binom{n-m}{m}$

Algebraic proof

\begin{align} \sum_k (-1)^k\binom{n-k}{n-m}\binom{m}{k} &=(-1)^m\sum_k (-1)^{m-k}\binom{n-k}{m-k}\binom{m}{k} \\&\stackrel{\text{n}}=(-1)^m\sum_k \binom{m-n-1}{m-k}\binom{m}{k} \\&\stackrel{\text{V}}=(-1)^m\binom{2m-n-1}{m} \\&\stackrel{\text{n}}=\binom{n-m}m \end{align} $\stackrel{\text{n}}=$ is the negation theorem.
$\stackrel{\text{V}}=$ is the (Chu)-Vandermonde identity.

Combinatorial proof sketch

Both sides answer the question

How many subsets of $\{1,2,\dots,n\}$ have size $m$, no two consecutive elements, and do not contain $n$?

The RHS is true because $$ (i_1,i_2,\dots,i_m)\implies (i_1,i_2+1,i_3+2,\dots,i_m+{m-1}) $$ is a bijection from increasing sequences of length $m$ whose entries are between $1$ and $n-m$, to subsets of $\{1,2,\dots,n-1\}$ with no adjacent elements (written in increasing order).

We count such subsets using the principle of inclusion exclusion. Namely, for $1\le i\le n-1$ let $E_i$ be the set of subsets where the $i^{th}$ smallest element is adjacent to the $(i+1)^{st}$ smallest element, and let $E_n$ be the set of subsets which include $n$. These are the bad subsets, so we want $|E_1^c\cap E_2^c\cap \dots \cap E_n^c|$, which is $$ \sum_{S\subseteq \{1,2,\dots,m\}} (-1)^{|S|}\left|\bigcap_{i\in S}E_i\right| $$ It turns out that $\left|\bigcap_{i\in S}E_i\right|=\binom{n-k}{m-k}$ whenever $|S|=k$. Roughly, when $k$ of the elements of the subset are constrained to be adjacent, we can group these elements together into a single block, so there are only $m-k$ elements to be chosen in a smaller space of $n-k$. I leave the messy details to you.