Find the remainder of the Euclidean division of $2018^{2019^{2020}}$ by 11.
We know that $2018^{10} \equiv 1(\bmod 11)$, so we seek to calculate the value of $2019^{2020}$ modulo

10. Since $2019 \equiv -1(\bmod 10)$ and 2020 is even, we have $2019^{2020} \equiv 1(\bmod 10)$. Therefore, $2018^{2019^{2020}} \equiv 2018^{1} \equiv 5(\bmod 11)$.