TITLE: How to prove that $\sum\limits_{i=0}^p (-1)^{p-i} {p \choose i} i^j$ is $0$ for $j < p$ and $p!$ for $j = p$
QUESTION [11 upvotes]: Let $p \in \mathbf{N}$. I don't know how to prove that
$$\sum_{i=0}^p (-1)^{p-i} {p \choose i} i^j=0 \textrm{ for } j \in \{0,\ldots,p-1\},$$ 
and 
$$\sum_{i=0}^p (-1)^{p-i} {p \choose i} i^p=p!$$ 
(Maybe using the following hint, but not necessarily.
Let's consider  Cramer's system of equations (where $0^0:=1$):
$$0^j x_0+1^j x_1+\ldots + p^j x_p=p! \delta_{j,p}$$
($j=0, \ldots,p$). The solution is probably given by $x_i=(-1)^{p-i} {p \choose i}$ for $i=0,\ldots, p$.)
Thanks.

REPLY [4 votes]: Note that $\displaystyle\binom{p}{i}\binom{i}{k}=\binom{p}{k}\binom{p-k}{i-k}$, Therefore,
$$
\begin{align}
\sum_i(-1)^i\binom{p}{i}\binom{i}{k}
&=\sum_i(-1)^i\binom{p}{k}\binom{p-k}{i-k}\\
&=\binom{p}{k}(1-1)^{p-k}\\
&=\binom{p}{k}0^{p-k}\tag{1}
\end{align}
$$
Which is $1$ when $k=p$ and $0$ when $k<p$. Thus, $\displaystyle T_pf=\sum_i(-1)^i\binom{p}{i}f(i)$ kills all combinatorial polynomials with degree less than $p$.
$\displaystyle k!\binom{i}{k}$ is a degree $k$ polynomial in $i$ with lead coefficient $1$. Thus, we can write
$$
i^j=j!\binom{i}{j}+\sum_{k=0}^{j-1}a_{jk} k!\binom{i}{k}\tag{2}
$$
for some integers $a_{jk}$. Therefore, putting together $(1)$ and $(2)$,
$$
\begin{align}
\sum_i(-1)^i\binom{p}{i}i^j
&=\sum_i(-1)^i\binom{p}{i}\left(j!\binom{i}{j}+\sum_{k=0}^{j-1}a_k k!\binom{i}{k}\right)\\
&=j!\binom{p}{j}0^{p-j}+\sum_{k=0}^{j-1}a_kk!\binom{p}{k}0^{p-k}\\
&=\left\{\begin{array}{rl}0&\text{if }j<p\\p!&\text{if }j=p\end{array}\right.\tag{3}
\end{align}
$$