A sum of products of binomial coefficients I am looking for a proof or a reference for the following
( apparent ) combinatorial identity:
$$
\sum_{i = s}^{s+t}\left(\,{-1}\,\right)^{\, i}{i \choose s}
{s \choose i - t}
=
\left(\,{-1}\,\right)^{s + t},\quad\mbox{where}\ s\geq t\geq 0\
\mbox{are integers}
$$
Any help will be appreciated.

We seek to verify that
$$\sum_{q=s}^{s+t} (-1)^q {q\choose s} {s\choose q-t}
= (-1)^{s+t}$$
where $s\ge t\ge 0.$ The LHS is
$$\sum_{q=0}^{t} (-1)^{s+q} {s+q\choose s} {s\choose s+q-t}
\\ = \sum_{q=0}^{t} (-1)^{s+q} {s+q\choose s} {s\choose t-q}
\\ = (-1)^s [z^t] (1+z)^s
\sum_{q=0}^{t} (-1)^{q} {s+q\choose s} z^q.$$
The coefficient extractor controls the range of the sum and we may write
$$(-1)^s [z^t] (1+z)^s
\sum_{q\ge 0} (-1)^{q} {s+q\choose s} z^q
\\ = (-1)^s [z^t] (1+z)^s
\frac{1}{(1+z)^{s+1}}
\\ = (-1)^s [z^t] \frac{1}{1+z} 
= (-1)^s (-1)^t = (-1)^{s+t}.$$
This is the claim.
