<s>[INST] 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 $(2015)(32) \mod 100 =80$?
 [/INST] If $X \equiv \sum_{k=1}^{2015} 2^{\left(5^k\right)} \pmod {100}$, then you can compute:
$$X_4 \equiv \sum_{k=1}^{2015} 2^{\left(5^k\right)} \pmod 4$$
$$X_{25} \equiv \sum_{k=1}^{2015} 2^{\left(5^k\right)} \pmod {25}$$
$X_4 \equiv 0$ in each term.  $2^{20} \equiv 1 \pmod {25}$, this you seem to have noted.  So
$$X_{25} \equiv \sum_{k=1}^{2015} 2^{\left(5^k \text{ mod } 20\right)} \pmod {25}$$
$5^k \equiv 5 \pmod 20$ for $k \ge 1$, so 
$$X_{25} \equiv \sum_{k=1}^{2015} 2^5 \equiv 2015\cdot 32 \equiv 5\pmod {25}$$
And $$\begin{cases} X \equiv 5 \pmod {25} \\ X\equiv 0 \pmod 4 \end{cases}$$ only has 1 solution $\pmod {100}$.
</s>