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.
 A: We have:
$$5^{2^{k+1}} - 1 = \left(5^{2^k} - 1\right)\left(5^{2^k}+1\right)$$
By induction the first factor is divisible by $2^{k+1}$, while the second one is obviously even and hence $5^{2^{k+1}} - 1$ is divisible by $2^{k+2}$
A: First assume that $5^{2^k} - 1$ is divisible by $2^{k + 1}$ so $2^{k + 1}\,|\,5^{2^k} - 1$. Now consider
$$5^{2^{k + 1}} - 1 = 5^{2\cdot 2^k} - 1 = (5^{2^k})^2 - 1 = (5^{2^k} - 1)(5^{2^k} + 1)$$
Now we need to show that $2^{(k + 1) + 1} = 2^{k + 2}$ divides $(5^{2^k} - 1)(5^{2^k} + 1)$. Our inductive assumption allows to know that $2^{k + 1}\,|\, 5^{2^k} - 1$. Now we just need to show that $2 \,|\, 5^{2^k} + 1$. Well $5^{2^k}$ will always be an odd number, so adding one makes it even and hence $2 \,|\, 5^{2^k} + 1$. 
Since $2$ and $2^{k + 1}$ divide $5^{2^k} + 1$ and $5^{2^k} - 1$, then $2\cdot 2^{k + 1} = 2^{k + 2} \,| 5^{2^{k + 1}} - 1\,$.
By induction then $2^{n+ 1}\,|\, 5^{2^n} - 1$ for all $n \in \mathbb{N}$.
A: $$5^{2^{k+1}}=(5^{2^k})^2=(a2^{k+1}+1)^2=1+2^{k+2}a(?)$$
A: Consider $n=k+1$:
$$5^{2^{k+1}}-1=\left(5^{2^k}\right)^2-1=(5^{2^k}-1)(5^{2^k}+1)=\cdots$$
A: 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:
$$(k2^{n+1})2p= (kp)2^{n+2}$$
