What is the last digit of this expression: $4^{2^1} + 4^{2^2} + 4^{2^3} + ...+ 4^{2^{2013}}$ The expression:
$4^{2^1} + 4^{2^2} + 4^{2^3} + ...+ 4^{2^{2013}}$
I know that the last digit of each of the sumands  is 6, but I have trouble proving that. I tried to prove it using induction, but then I realized that I don't have a clue how to write it properly. 
Also, because of the fact that the last digit of every sumand is 6, the last digit of the given expression is the last digit of $2013 * 6$ which is 8.

$4^2\equiv 6 \mod 10$ and $6^2\equiv 6\mod 10$, which is equivalent to $4(^2)^2=4^4$.
With the $(6 \mod 10)$ relation, we can continue doubling the exponent of $4$ with ease.
$(4^4)^2=4^8\equiv 6^2=6 \mod 10, \dots$
