How to prove that $a^{2^n} - 1$ is divisible by $4\cdot2^n$ by induction

Question

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.

Answer 1

Hint:

$$\frac{a^{2^{n+1}}-1}{a^{2^n}-1}=a^{2^n}+1$$ which is even.

Answer 2

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.