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.

Sorry if this is not rigorous enough, but I'll give it a try: consider 2^4=16 which ends in a 6. We know that when multiplying two numbers, to find the last digit of the product we only have to multiply the last digits of the 2 terms we are multiplying. So since 6*6=36, 2^4n  (for n>=1) will have a last digit of 6. So our power of 2 needs to be a multiple of 4. Well any positive integer power of 2 greater than 1 will be divisible by 4, since 2^n/2^2 =2^(n-2), which is an integer. So we have concluded that 2 to the any power of 2 will have it's last digit as 6, and from that it is easy to conclude that if you add 1 it will always end in 7.  
