Alternate proof for $2^{2^n}+1$ ends with 7, n>1. I have a proof by induction that $2^{2^n}+1$ end with 7. I've been trying to prove that within the theory of rings and ideals, but haven't achieved it yet. The statement is equivalent to $2^{2^n}-6$ ends with zero, so

Prove that for $$ e \in \mathbb{N} : e=2^n \Rightarrow 2^e-6 \in (10)\subset\mathbb{Z}$$ 

I'm not sure if this is the easiest equivalent statement to prove this in the language of rings. any help?
or alternatively $$ e\in \bar{0}\in\mathbb{Z}_4 \Rightarrow 2^e-6 \in \bar{0} \in \mathbb{Z}_{10} $$
is also proven by induction.

Here is an alternate proof, it is basically the same idea as TA or Marvis, but probably presented in a "elementary" way.
For $n \geq 2$ then
$$2^{2^n}-16=16^{2^{n-2}}-16=16[16^{\alpha}-1]=16(16-1)(\mbox{junk})$$
Since 16 is even and 15 is divisible by 5, it follows that $2^{2^n}-16$ is a multiple of 10....
