What is the remainder of the Euclidean division of $2022^{2022^{2022}}$ by 11?
By Fermat's Little Theorem, we have \(2022^{10} \equiv 1 \pmod{11}\). Thus, if we know the value of \(2022^{2022} \mod 10\), we would be very happy.

We look at the powers of 2 modulo 10: they have a period of 4: 2, 4, 6, 8, 2, ... Since 2022 is congruent to \(2 \pmod{4}\), we get \(2022^{2022} \equiv 2^{2022} \equiv 4 \pmod{10}\).

Moreover, \(2024 = 184 \times 11\), so \(2022 \equiv -2 \pmod{11}\).

Finally, \(2022^{2022^{2022}} \equiv 2022^{10k} \times (-2)^{4} \equiv 16 \equiv 7 \pmod{11}\).