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.
 A: The hint of @darijgrinberg is valuable and deserves an answer by its own. 

We obtain
  \begin{align*}
\color{blue}{\sum_{q=s}^{s+t}}&\color{blue}{(-1)^q\binom{q}{s}\binom{s}{q-t}}\\
&=\sum_{q=0}^t(-1)^{q+s}\binom{q+s}{q}\binom{s}{q+s-t}\tag{1}\\
&=(-1)^s\sum_{q=0}^t\binom{-s-1}{q}\binom{s}{t-q}\tag{2}\\
&=(-1)^s\binom{-1}{t}\tag{3}\\
&=(-1)^s\frac{(-1)(-2)\cdots(-t)}{t!}\\
&\,\,\color{blue}{=(-1)^{s+t}}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we shift the index to start with $q=0$.

*In (2) we use the binomial identities $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ and $\binom{p}{q}=\binom{p}{p-q}$.

*In (3) we apply the Chu-Vandermonde identity.
A: 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.
