Proving an interesting (falling) factorial identity

Question

Let $(x)_{n}=x(x-1)\ldots(x-n+1)$ for $n>0$ and $(x)_0=1$.

Prove that if $n<k$, then \begin{equation} \displaystyle\sum_{i=0}^k (-1)^i {{k}\choose{i}}(x-i)_n=0 \end{equation}

Any thoughts? Note that $n=0$ is trivial since you are summing alternating binomial coefficients.

Answer 1

For example:

$\displaystyle \sum\limits_{n=0}^\infty \frac{z^n}{n!} \sum\limits_{i=0}^k (-1)^i {\binom k i} (x-i)_n = \sum\limits_{i=0}^k (-1)^i {\binom k i} \sum\limits_{n=0}^\infty \frac{z^n}{n!} (x-i)_n =$

$\displaystyle = \sum\limits_{i=0}^k (-1)^i {\binom k i} (1+z)^{x-i} = z^k (1+z)^{x-k} = \sum\limits_{j=0}^\infty {\binom {x-k} j} z^{k+j} $

Comparing the coefficients of $\,z^n\,$ proofs the claim.