Recall that $a$ is a $k$-th power iff $a^\frac{p-1}{\gcd(p-1,k)} \equiv 1 \mod p$. We show that $n$ is a fourth power $\mod p$: We have $16n \equiv -4 \equiv (2i)^2 \mod p$, thus we want that $2i$ is a square $\mod p$ (here $i$ is such that $i^2 \equiv -1 \mod p$, existing by $p \equiv 1 \mod 4$). - If $p \equiv 1 \mod 8$, then $2$ is a quadratic residue and $i$ is, too, the latter by $i^{p-1}2 = (-1)^\frac{p-1}4 = 1 \mod p$. - If $p \equiv 5 \mod 8$, then $2$ is nonsquare $\mod p$ and $i$ is, too, the latter by $i^\frac{p-1}2 \equiv (-1)^\frac{p-1}4 = -1 \not\equiv 1 \mod p$. In both cases, $2i$ is a quadratic residue, thus there is an $a$ with $a^4 \equiv n \mod p$. Or shorter: $(1+i)^4 \equiv (2i)^2 \equiv -4 \mod p$, giving the same result with $a=1+i$. This solves directly a. (and shows more), but also b. follows: $n^n \equiv (a^4)^\frac{p-1}4 = a^{p-1} \equiv 1 \mod p$.