Consider the following binomial summation:
$$\sum_{k=0}^{m}\binom{k}{n}\left(-1\right)^k$$
For $m\ge n\ge0$ and $m,n \in \mathbb N$
This is indeed the alternating sign Hockey-stick identity.
$$\sum_{k=0}^{m}\binom{k}{n}\left(-1\right)^k=\sum_{k=n}^{m}\binom{k}{k-n}\left(-1\right)^k=\left(-1\right)^{-n}\sum_{k=n}^{m}\binom{-n-1}{k-n}$$
Setting $k-n \mapsto k$ follows:
$$=\left(-1\right)^{-n}\sum_{k=0}^{m-n}\binom{-n-1}{k}$$
I also tried the classical method:
$$S_{n}:=\sum_{k=n}^{m}\binom{k}{n}\left(-1\right)^k=\sum_{k=n}^{m}\binom{k+1}{n+1}\left(-1\right)^k-\sum_{k=n}^{m}\binom{k}{n+1}\left(-1\right)^k$$
Setting $k+1 \mapsto k$ yields:
$$=-\sum_{k=n+1}^{m+1}\binom{k}{n+1}\left(-1\right)^k-\sum_{k=n}^{m}\binom{k}{n+1}\left(-1\right)^k$$ $$=-\sum_{k=n+1}^{m}\binom{k}{n+1}\left(-1\right)^k-\sum_{k=n+1}^{m}\binom{k}{n+1}\left(-1\right)^k+\binom{m}{n}\left(-1\right)^m=-2S_{n+1}+\binom{m}{n}\left(-1\right)^m$$
Using this , a recurrence relation can be derived:
$$S_{n+1}=\frac{\binom{m}{n}\left(-1\right)^m-S_n}{2}$$
But it's not what I want!
So how we can find a closed form for alternating sign Hockey-stick identity?