elementary number theory - Prove that $5^{2^n} - 1$ is divisible by $2^{n+1}$ for all $n ≥ 1$

I made the following using induction:

If $n=1$ then the proposition is true: $5^{2^1} - 1=24$ is divisible by $2^{1+1} = 4$

Now I suppose that for a natural number $k$, $5^{2^k} - 1$ is divisible by $2^{k+1}$ is true. And I want to prove (using this) that the proposition is true for $n=k+1$ but I don't know how to do this.

I appreciate the help you give me.

Answer

Step $n+1$:

$$\begin{align} 5^{2^{n+1}}-1 &= 5^{2^n \times 2}-1 \\ &= (5^{2^n})^2-1 \\ &= (5^{2^n}-1)(5^{2^n}+1) \\ &=(k2^{n+1})(5^{2^n}+1) \end{align}$$

For the first factor above the hypothesis (step $n$) is used,

the second factor is even, say $2p$, since a power of $5$ is odd.

Combining:

$$(k2^{n+1})2p= (kp)2^{n+2}$$