Proof for $\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$ I am stuck at the following exercise:
Prove that
$\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$.
I tried to prove the Expression by induction but I cannot find a way to prove the implication
$$7\mid(1 + 2^{2^n} + 2^{2^{n+1}}) \Rightarrow 7\mid(1 + 2^{2^{n+1}} + 2^{2^{(n+1)+1}}).$$
Any help would be appreciated very much.

Induction will work here.  Assume $1+2^{2^n} + 2^{2^{n+1}}$ is divisible by 7.
Note that $2^{2^{n+2}} = 2^{2^n4} = 16^{2^n} = (14+2)^{2^n} = 14M+2^{2^n}$ for some integer $M$.  Then $1+2^{2^{n+1}} + 2^{2^{n+2}} = 1+2^{2^{n+1}}+(14+2)^{2^n} = 1+2^{2^{n+1}} +2^{2^n}+14M$, which must be a multiple of 7, since the first three terms are from the induction hypothesis.
