Cycling twos

Note the last digit of powers of 2:

$2^{1} \rightarrow 2\\ 2^{2} \rightarrow 4\\ 2^{3} \rightarrow 8\\ 2^{4} \rightarrow 6$

This pattern will then loop around

$2^{5} \rightarrow 2\\ 2^{6} \rightarrow 4\\ ...$

This means that $2^{n}$ will end in either $2,4,6,8$ depending on $n$ 's value in $\mod 4$ .

$2^{2k}=4^{k}$

so $2$ raised to an even power will result in congruence to $0\mod 4$ .

$2$ raised to any power is even, so our expression is:

$2^{4k} \equiv 0 \mod 4$

Looking at our cycle, $n \equiv 0 \mod 4$ , so our number ends in $\boxed{6}$ .

The given number is $2^{2^{16}}=2^{(4*2^{14})}=(2^4)^{2^{14}}=16^{2^{14}} \equiv 6^{2^{14}}\equiv 6 \pmod{10}$ since any power of 6 is congruent to 6.

To find last digit, find remainder of Index divided by 4,

If remainder is 1 then unit digit is base^1

remainder is 2 then unit digit is base^2

remainder is 3 then unit digit is base^3

remainder is 0 then unit digit is base^4

In above case it is 2^256 so 256 divided by 4 remainder is 4 so

unit digit is unit digit of 2^4 = 6

2^1=2; 2^2=4; 2^3=8; 2^4=16; 2^5=32.......so on! so every 4th power consists last digit 6. the given number 2^2^2^2^2 can be express as 2^256 which is 2s 4th power n last digit is 6!