Let $p$ be an odd prime such that $p \not \mid n$. Show that the multiplicative inverse of $n^{\frac{p-1}{2}} \pmod p$ is $n^{\frac{p-1}{2}} \pmod p$.
Let's set $a = n^{\frac{p-1}{2}}$ and $b = n^{\frac{p-1}{2}}$ modulo $p$, and then consider $ab \pmod p$
$$ \begin{align} ab & \equiv n^{\frac{p-1}{2}} \times n^{\frac{p-1}{2}} \pmod p \\ \\ & \equiv n^{p-1} \pmod p \\ \\ & \equiv 1 \pmod p \text { using FlT} \end{align}$$
This means $b$ is $a^{-1}$, that is, $n^{\frac{p-1}{2}}$ is its own multiplicative inverse modulo $p$.