How to prove that $a^{2^n} - 1$ is divisible by $4\cdot2^n$ by induction Given that $a$ is any odd number and $n$ is any integer
I got to $(2k + 3)^{2^k}\cdot(2k + 3)^{2^k} - 1$ at the $(k+1)$-th step.

For $n=1$ this is obvious, because $a^2-1=(a-1)(a+1)$, and both factors are even, so $2^2 = 4\cdot 2^0$ divides $a^2-1$.
Assume this is true for $n$, so $4\cdot 2^n$ divides $a^{2^n}-1$. Then
\begin{align*}
a^{2^{n+1}}-1 & = a^{2^n\cdot2}-1 \\
& = (a^{2^n}-1)(a^{2^n}+1) \\
& = 4\cdot 2^n \cdot k\cdot(a^{2^n}+1),
\end{align*}
for $k$ an integer. But now $a^{2^n}$ is odd, so $a^{2^n}+1$ is even, hence $a^{2^n}+1 = 2h$ for $h$ an integer. Therefore
$$a^{2^{n+1}}-1 = 4\cdot 2^n \cdot k \cdot 2h = 4\cdot 2^{n+1} \cdot hk$$
and the claim follows.
