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.
 A: From little Fermat, we have $$2^{4k+3} \equiv 3 \pmod{5}$$
We have $4 \vert 2^n$ for $n \geq 2$. Hence, $2^n-1 \equiv 3 \pmod4$. Hence, we get
$$2^{2^n-1} \equiv 3 \pmod{5}$$
Hence,
$$2^{2^n-1} = 5k+3 \implies 2^{2^n} = 10k + 6 \implies 2^{2^n}+1 = 10k+7$$
A: Not true if $n=1$. True for $n>1$. 
Hint: If $K$ ends in $6$ then so does $K^2$.
Now note that $\left(2^{2^n}\right)^2 = 2^{2^{n+1}}$.
A: 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....
A: 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.  
