Evaluating $\sum_{i=0}^k \frac{(-1)^{k-i}k! (n+i)^{k-1}}{i!(k-i)!}$ How to evaluate the the following sum where $n $ is an integer greater than $0$.
$$\sum_{i=0}^k \frac{(-1)^{k-i}k! (n+i)^{k-1}}{i!(k-i)!}$$
I think the answer is $0$, but I can not prove it.

The statement is true, one way to see it is to define a operator $L$ on the spaces of real sequences indexed by $\mathbb{N}$ by:
$$a = (a_k)_{k\in\mathbb{N}} \quad\mapsto\quad L(a)_k = a_{k+1},\;k \in \mathbb{N} $$
i.e. shift the index of the sequence $a$ to the left by one unit. In terms of $L$, the inequality can be formally rewritten as
$$
\sum_{i=0}^k (-1)^{k-i} \binom{k}{i} (n+i)^{k-1} =
\sum_{i=0}^k (-1)^{k-i} \binom{k}{i} L^i(n^{k-1}) = (L - 1)^k(n^{k-1}) = 0$$
This is equivalent to claiming: 

Given $n^{k-1}$, a polynomial in $n$ of degree $k-1$, the $k^{th}$ time finite difference of it vanishes.

To prove the statement, it just suffices to observe for any polynomial $P(n)$ of degree $m$,
the finite difference of $P(n)$, $P(n+1) - P(n)$, is a polynomial of degree $m-1$.
