Congruence Modulo Powers of Primes I need help with the following:
Let $p$ be an odd prime and $x, y \in \mathbb{Z}$. Then $x \equiv y$ mod $p^k \implies x^p \equiv y^p$ mod $p^{k+1}$.
I know that I can write $x^p - y^p$ as $(x - y)F(x,y)$ for some ugly polynomial $F$, but I'm not sure whether that's too much help.
Ta

Why not directly?:
$$x\equiv y\pmod{p^k}\implies x=y+mp^k\;,\;\;m\in\Bbb Z\implies$$
$$x^p=\sum_{n=0}^p\binom pny^n(mp^k)^{p-n}\;\;(**)$$
But all the binomial coefficients except the first one and the last one are dividisible by $\;p\;$ (either you know this or you can prove it by induction...) , so we get:
$$(**)=(\text{something divisible by}\;p^{k+1})+m^pp^{pk}+y^p=y^p\pmod{p^{k+1}}$$
since, of course, $\;pk>k+1\;$
