Prove that $a^{2^n}=1 \mod 2^{n+2}$
Solution 1:
$$a^{2^n}-1=(a-1)(a+1)(a^{2}+1)\cdots(a^{2^{n-1}}+1)$$
If $a$ is odd then each term in the factorisation is even (i.e. divisible by $2$), there are $n+1$ such terms hence $a^{2^n}-1$ is divisible by $2^{n+1}$. Since either $a-1$ or $a+1$ is a multiple of $4$, we have an additional factor of $2$, therefore $a^{2^n}-1$ is divisible by $2^{n+2}$