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Since $p:=4n+1$ is a prime number, each non-zero remainder modulo $p$ possesses a unique multiplicative inverse. Since $-4\cdot n\equiv 1(\mathrm{mod}p)$, we have $n\equiv (-4{)}^{-1}(\mathrm{mod}p)$, from which we deduce that $n\equiv -(2^{-1}{)}^2$. Consequently, $$n^{2n}-1\equiv (-(2^{-1}{)}^2{)}^{2n}-1\equiv (2^{-1}{)}^{4n}-1\equiv 0(\mathrm{mod}p),$$ by Fermat's Little Theorem.