Gotta find that remainder

Using the facts that $2018 \equiv 2 \pmod{7}$ and $2^3 \equiv 1 \pmod{7}$ we see that

$\begin{aligned} 2018^{8012} &\equiv 2^{8012} \pmod{7} \\ \\ &\equiv (2^3)^{2670} \cdot 2^2 \pmod{7} \\ \\ &\equiv 1^{2670} \cdot 2^2 \pmod{7} \\ \\ &\equiv 4 \pmod{7} \end{aligned}$