Is it true that $2^{2^{n}}$ always ends in the digit $6$? As the question says, is it true, that for $n >1$, 
$2^{2^{n}}$ ends in the digit 6?
How would one prove this? It seemed true.
I considered writing it as
$$2^{2^{n}} = \prod_{k=0}^{n} 2^{ n\choose k }$$ but I'm not sure that's useful. Any help, ideas, or even hints?
 A: It's more general than that: the last digit of the decimal expansion of every number of the form $2^{4n}$ is $6$. That's because $2^{4n}=16^n$ and, of course, if the last digit of a number is $6$, then the last digit of any of its powers is also $6$ (since $6\times6=36$).
A: You can use induction. Assume that $2^{2^n}$ ends with 6. Now $2^{2^{n+1}}  = 2^{2^n + 2^n}= 2^{2^n} . 2^{2^n}$ and the multiplication of two numbers ending with 6 must end with 6.
A: $2^{2^n}\equiv0\pmod2$ for $n\ge0$
Now $2^{2^n}=(2^2)^{2^{n-1}}\equiv(-1)^{2^{n-1}}\pmod5$
For $n-1\ge1,(-1)^{2^{n-1}}\equiv1\pmod5$
So, for $n\ge2, 2^{2^n}\equiv1+5\pmod{10}$
A: Using modular arithmetic would be easiest. For a number $k$ to end in the digit 6 is the same as $k\equiv 6\mod 10$, which, by the Chinese Remainder Theorem is the same as $$\left\{\begin{array}{r}k\equiv0\mod 2 \\k\equiv 1\mod 5\end{array}\right.$$
The top one is easy. The bottom is also not too difficult if you note that $2^4\equiv 1\mod 5$. Can you take it from here?
