TITLE: Fermat's Little Theorem: exponents powers of p
QUESTION [9 upvotes]: I was working with congruence classes and encountered Fermat's little theorem:
$$a^{p } \equiv a \mod p$$
But I noticed that a$^{p^{k}} \equiv a \mod p$. 
I used induction on $k$ but I'm still not convinced. Can anyone give an intuitive way to see why this is?

REPLY [2 votes]: When I encountered the same theorem I used induction on a: $(a+1)^{p} = a^{p} + p(\mathrm{something}) + 1 ≡ a + 1$
Since $p$ is prime, all $\binom {k} {p}$ are divisible by $p$.