Prove by induction that $1+2^{2^n}+2^{2^{n+1}}$ is divisible by $7$ As said in the title you have to show by induction that 1+$2^{2^n}+2^{2^{n+1}}$ is divisible by 7. So you start with n=0, that gives 1+2+4=7. So the start is shown.
Let 1+$2^{2^n}+2^{2^{n+1}}$ be divisble by 7 for a n.
I've tried several attempts but I ended up in a mess.
For example the latest attempt:

We have $1+2^{2^{n+1}}+2^{2^{n+2}}$
Adding $2^{2^n}-2^{2^n}$ gives you
$1+2^{2^{n+1}}+2^{2^n}+2^{2^{n+2}}-2^{2^n}$
As the first three summands resemble our Assumption, only $2^{2^{n+2}}-2^{2^n}$ needs to be proved as a multiple of seven.

But I am stuck at trying to show this.
As stated that might be a incorrect approach by myself, so I'm not really sure whether or not that is going in the correct direction. In the end I'd really appreciate some help on this question!

Starting with $n = 0$, we have $1+2^{2^0}+2^{2^{0+1}} = 7$, which is clearly divisible by $7$. Now suppose inductively that $n \ge 1$ and argument holds for all $n$. Then for $n+1$, we have
$$1+2^{2^{n+1}}+2^{2^{n+2}} = 1+2^{2\cdot2^n}+2^{2\cdot2^{n+1}} = 1+(2^{2^n})^2+(2^{2^{n+1}})^2$$
Now and and subtract $(2^{2^n})^2$, we have
$$1+2(2^{2^n})^2+(2^{2^{n+1}})^2-(2^{2^n})^2 = [1+(2^{2^n})^2]^2-(2^{2^n})^2$$ $$ = [1+(2^{2^n})^2+2^{2^n}]\cdot[1+(2^{2^n})^2-2^{2^n}]$$
$$ = [1+2^{2^{n+1}}+2^{2^n}]\cdot[1+2^{2^{n+1}}-2^{2^n}]$$
Here, notice that by inductive hyphothesis, $7|(1+2^{2^{n+1}}+2^{2^n})$. Therefore $7$ divides the whole expression and argument holds for $n+1$. Therefore by induction, argument holds for all $n$.
