Fermat's Little Theorem: exponents powers of p 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?
 A: If $a$ is divisible by $p$, it is obvious.
If not, Fermat's Little Theorem is equivalent to $a^{p-1}\equiv 1\bmod p$.
Raising both sides to any power shows that $a^x\equiv 1\bmod p$ for any $x$ a multiple of $p-1$.
$p^k-1$ is a multiple of $p-1$: $(p-1)(p^{k-1}+p^{k-2}+\ldots+1)$.
A: Hint: Take $k = 2$. We have $$a^{p^2} \equiv (a^p)^p \equiv  (a)^p \equiv a \pmod p$$
A: Hint $\ $ By an obvious induction, any set closed under multiplication is closed under powers.
Note that $\ \color{#c00}{a^J} \equiv a\equiv \color{#0a0}{a^K}\,\Rightarrow\, a^{\color{#c00}J\color{#0a0}K}\equiv (\color{#c00}{a^J})^{\color{#0a0}K}\equiv \color{#0a0}{a^K}\equiv a\,$ so the set $\,S\,$ of $\,N\,$ with $\,a^N\equiv a$ 
satsifies $\ \ \color{#c00}J,\color{#0a0}K\in S\ \Rightarrow\ \color{#c00}J\color{#0a0}K\in S,\ $ i.e. $S\,$ is closed under multiplication, so under powers.
In particular $\ a^P\equiv a\,\Rightarrow\, p\in S\,\Rightarrow\, p^k\in S\,$ for all $\,k\ge 1$.
Remark $\ $ Note how bringing to the fore the innate monoid structure (closure under product) serves to reduce the induction to a trivial induction, that monoids are closed under powering. Such simplifications are often possible in common inductive proofs, so it is worthwhile to first search for such simplifying structure before diving head-first into brute-force inductions.
A: $$a^p = a(modp) ==> a^(p^2) = a^p = a(modp) ==> a^(p^3) = a^p = a(modp) ==> a^(p^k) = a^p = a(modp). $$
A: $a^{p^k}=(((a^p)^p)...) \equiv ... \equiv( (a^p)^p)^p \equiv (a^p)^p \equiv a^p \equiv a$ by simply iterating Fermat's Little Theorem within the delimeters.
A: 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 $p$.
