I need help with the following problem, I don't know how to continue. Let $p$ be a prime. Prove that if $p \mid a-b$ then: $$p^{n+1} \mid a^{p^n}-b^{p^n}$$
At first I thougt the following:
$$p \mid a-b$$ $$a \equiv b\ (p)$$ $$a^{p^n} \equiv b^{p^n}\ (p)$$ But I don't know how to go from $\bmod p$ to $\bmod p^{n+1}$, or if there's a simpler way to prove it. Could you provide me with any hint or another way of thinking of this problem? Thank you.
You can prove this by induction. The main step is the following
Lemma: If $a \equiv b \mod rq$ for some $q, \, r$, then $a^q \equiv b^q \mod rq^2$.
Proof: Suppose $a \equiv b \mod rq$, that is $a = b + krq$ for some integer $k$. Then by the binomial theorem $$ a^q = b^q + q b^{q-1}\cdot krq + + \sum_{j = 2}^q \binom{q}{j} b^{q-j} (krq)^j = b^q +\ell r q^2 $$ for some integer $\ell$. Therefore $a^q \equiv b^q \mod rq^2$.
Now to prove the statement, use induction. The base step $n = 1$ is done by using the Lemma with $q = p$ and $r = 1$.
The induction step works as follows: Assume $a^{p^n} \equiv b^{p^n} \mod p^{n+1}$. Thus the assumption of the Lemma holds with $q = p, r = p^n$. Apply the Lemma and obtain $$ a^{p^{n+1}} = \left(a^{p^n}\right)^p \equiv \left(b^{p^n}\right)^p = b^{p^{n+1}} \mod rp^2 = p^{n+2} \, . $$ The proof is complete.