Balti Tee 2018 · Ülesanne 18

Since $p := 4n + 1$ is a prime number, each non-zero remainder modulo $p$ possesses a unique multiplicative inverse. Since $-4 \cdot n \equiv 1 \pmod{p}$, we have $n \equiv (-4)^{-1} \pmod{p}$, from which we deduce that $n \equiv -(2^{-1})^2$. Consequently,

$$n^{2n} - 1 \equiv (-(2^{-1})^2)^{2n} - 1 \equiv (2^{-1})^{4n} - 1 \equiv 0 \pmod{p},$$

by Fermat's Little Theorem.