proof that the binomial sum is equal to 1 I'm trying to prove the following identity: 
let $k$ and $s$ be positive integers and let $k\ge s\ge 1$
$$\sum_{i=0}^{k-s} (-1)^i{s-1+i \choose s-1}{k \choose s+i} = 1$$
I've tried to use a generating function method to prove this formula, but I don't see how to apply it here just yet. With all the other methods I got stuck. How can it be proven? Thanks in advance.
 A: I found it convenient to let $d=k-s$, so that 
$$\begin{align*}
\sum_{i=0}^{k-s} (-1)^i{s-1+i \choose s-1}{k \choose s+i}&=\sum_{i=0}^{k-s}(-1)^i\binom{s-1-i}i\binom{k}{k-s-i}\\
&=\sum_{i=0}^{d}(-1)^i\binom{s-1+i}i\binom{s+d}{d-i}\;.
\end{align*}$$
Let
$$f(x)=\sum_{n\ge 0}(-1)^n\binom{s-1+n}nx^n=\sum_{n\ge 0}\binom{s-1+n}n(-x)^n=\frac1{(1+x)^s}$$
and
$$g(x)=\sum_{n\ge 0}\binom{s+d}nx^n=(1+x)^{s+d}\;.$$
Then
$$\sum_{i=0}^{d}(-1)^i\binom{s-1+i}i\binom{s+d}{d-i}$$
is the coefficient of $x^d$ in 
$$f(x)g(x)=(1+x)^d\;,$$
which is clearly $1$.
A: Here is another variation based upon the usage of the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a generating series.

We obtain
  \begin{align*}
\sum_{i=0}^{k-s}&(-1)^i\binom{s-1+i}{s-1}\binom{k}{s+i}\\
&=\sum_{i=0}^{k-s}\binom{-s}{i}\binom{k}{s+i}\tag{1}\\
&=\sum_{i=0}^{\infty}[z^i](1+z)^{-s}[u^{s+i}](1+u)^k\tag{2}\\
&=[u^s](1+u)^k\sum_{i=0}^{\infty}u^{-i}[z^i](1+z)^{-s}\tag{3}\\
&=[u^s](1+u)^k\left(1+\frac{1}{u}\right)^{-s}\tag{4}\\
&=[u^s](1+u)^ku^{s}\left(u+1\right)^{-s}\\
&=[u^{0}](1+u)^{k-s}\\
&=1
\end{align*}

Comment:


*

*In (1) we use the binomial identity
\begin{align*}
\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k=\binom{n+k-1}{n-1}(-1)^k
\end{align*}

*In (2) we apply the coefficient of operator and set the upper limit of the sum to $\infty$ without changing anything, since we add only zeros.

*In (3) we do some rearrangements and use 
\begin{align*}
[z^{n+k}]A(z)=[z^n]z^{-k}A(z)
\end{align*}

*In (4) we use the relationship
\begin{align*}
A(z)=\sum_{j=0}^\infty a_jz^j=\sum_{j=0}^{\infty}\left([u^j]A(u)\right)z^j
\end{align*}
