Proving that $2^{n+2} \mid a^{2^{\large{n}}} - 1 $, for $a \in \mathbb{Z}$ odd, $n \in \mathbb{N}$ How can I prove this? I was thinking of using strong induction on $n$, and I've managed to prove the base case using induction again on $a$,
$$2^3 \mid a^{2} - 1$$
but I couldn't work my way around the rest of the proof with the usual properties of divisibility (if $m \mid n$, then $a^m - 1 \mid a^n - 1$, etc). 

Hint: 
$$
a^{2^n} - 1 = (a^{2^{n-1}} - 1)(a^{2^{n-1}} + 1)
$$
