three digits of $2^{8064}$ ?
Here's my method
I use two powerful hints :
-The first one consists to take the number $\pmod{1000}$ because we want the last three digits.
-The second one is a powerful theorem from Euler-Gauss which shows that if we have an integer of the form $p_1^{k_1}...p_l^{k_l}$ with $p_1,...,p_l$ prime numbers then for every integer $a$ we have $a^{({p_1}^{k_1}-{p_1}^{k_1-1})...{(p_l^{k_l}-p_l^{k_l-1})}}\equiv 1 \pmod{p_1^{k_1}...p_l^{k_l}}$ and $\gcd(p_1^{k_1}...p_l^{k_l},a)=1$.
Applying these two results I get : $1000=2^3\times5^3$ which means that $2^{4\times 100}\equiv 1 \pmod{1000}\Leftrightarrow 2^{400}\equiv 1 \pmod{1000}$. We find a period for the power of $2$. But the problem is that $\gcd(2,1000)\ne 1$. So it won't work with this theorem and its simple hypothesis.
Now using @Joffan's argument on Carmichael, we have the fact that the order of $2$ is $100$. So $2^{100}\equiv 1 \pmod{1000}$.
So $2^{100\times 80+64} \equiv 2^{64}\equiv (((2^8)^2)^2)^2 \equiv 616 \pmod{1000}$
However, excepting computing it is there a faster method ?
Thanks in advance !