Proof By Induction that $3^{(2^n)} -1$ is divisible by $2^{(n+2)}$

Question

How do I prove the $(n+1)$-th case for this equation?

Answer 1

$$3^{2^{n+1}}-1 = (3^{2^{n}}-1)(3^{2^{n}}+1) = k2^{n+2}(3^{2^{n}}+1)$$

Now, note that $$3^{2^{n}}+1$$ is obviously even ($3^k$ is odd) and therefore divisble by $2$, and therefore we get: $$3^{2^{n+1}}-1 = 2^{n+2}\cdot 2m = 2^{n+3} \cdot m$$

Which shows that $$2^{n+3} \mid 3^{2^{n+1}}-1$$