Last two digits of $2^5+2^{5^2} +\dots +2^{5^{2015}}$ 
Let $$N=2^5+2^{5^2} +\dots +2^{5^{2015}}.$$ Find the last two digits of $N$.

It suffices to find $N \mod 100$. And I observed that $2^{20+n}=2^n\pmod{100}$ for $n \ge 2$. So is the answer just $(2015)(32) \mod 100 =80$?

Notice that
$$
32^5=2^{25} \equiv 2^{10}\cdot 2^{10} \cdot 2^5 \equiv 24^2 \cdot 32 \equiv 32 \mod 100
$$
(I think what you wrote there does not make sense as stated)
and it follows that 
$$
2^{5^k} \equiv 32 \mod 100
$$
for all $k\ge 1$
So indeed
$$
2^{5}+2^{5^2}+\cdots+2^{5^{2015}} \equiv 32 \cdot 2015 \equiv 64480 \equiv 80  \mod 100
$$
