Use the PIE to prove an identity Use the PIE to prove,
$n!{{n-1}\choose {k-1}} = \sum_{i=1}^k (-1)^{k-i} {k \choose i} \bar i^{n}, \ 1 \le k \le n$
Where,
$\bar i^{n}= i(i+1)...(i+n-1)$
Edit
Actually,I can solve it now. And it's a cool question. So leaving it for my fellow friends out there.
 A: You’re right: it is a cute question. I’ll use the notation $i^{\overline n}$ for the rising factorial, however. First we note that
$$\begin{align*}
\sum_{i=1}^k(-1)^{k-i}\binom{k}ii^{\overline n}&=\sum_{i=1}^k(-1)^{k-i}\binom{k}i\frac{(n+i-1)!}{(i-1)!}\\\\
&=n!\sum_{i=1}^k(-1)^{k-i}\binom{k}i\binom{n+i-1}{i-1}\;,
\end{align*}$$
so the identity reduces to
$$\binom{n-1}{k-1}=\sum_{i=1}^k(-1)^{k-i}\binom{k}i\binom{n+i-1}{i-1}\;.\tag{1}$$
The lefthand side of $(1)$ is the number of ways of distributing $n$ indistinguishable objects amongst $k$ distinguishable boxes so that no box is empty. The binomial coefficient $\binom{n+i-1}{i-1}$ is the number of ways of distributing $n$ indistinguishable objects amongst $i$ distinguishable boxes, where the boxes are allowed to be empty. Thus, 
$$\binom{k}i\binom{n+i-1}{i-1}$$
is the number of ways of distributing $n$ indistinguishable objects amongst $k$ distinguishable boxes so that at least $k-i$ boxes are empty, and $(1)$ follows immediately by the inclusion-exclusion principle.
A: By way of enrichment I present an algebraic proof.
I upvoted @Brian M. Scotts's answer.
Suppose we are trying to prove
$${n-1\choose k-1}
= \frac{1}{n!} \sum_{q=0}^k (-1)^{k-q} {k\choose q} 
q^{\overline{n}}$$
where $1\le k\le n.$
This is
$${n-1\choose k-1}
= \sum_{q=0}^k (-1)^{k-q} {k\choose q} 
{q+n-1\choose n}.$$
Introduce the integral representation
$${q+n-1\choose n}
=\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{q+n-1}}{z^{n+1}} \; dz.$$
This gives for the sum the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1}}{z^{n+1}} 
\sum_{q=0}^k (-1)^{k-q} {k\choose q} (1+z)^q
\; dz$$
which simplifies to
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1}}{z^{n+1}} 
(1+z-1)^k
\; dz$$
or
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1}}{z^{n-k+1}} 
\; dz.$$
This can be evaluated by inspection to give
$${n-1\choose n-k} = {n-1\choose k-1}.$$
Of course when  $k\ge n+1$ the integrand turns into an entire function
giving the value zero.
