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.