If p is prime and $\gcd(a,p-1)=1$ then $x^{1/a}=x^{-a}$ for any $x \in \mathbb{Z}_p^*$ where $a^{-1}$ is the inverse of a in $\mathbb{Z}_{p-1}$

Question

I need to prove the following.

If p is prime and $\gcd(a,p-1)=1$ then $b^{1/a}=b^{-a}$ exists for any $b \in \mathbb{Z}_p^*$ where $a^{-1}$ is the inverse of $a$ in $\mathbb{Z}_{p-1}$

I know $\mathbb{Z}_p^*$ is a cyclic group with $p-1$ elements. I think that, as a result, each $b$ has a distinct root of any degree, but I don't know how to show that for sure and I'm not sure how $a^{-1}$ being the inverse of $a$ in $\mathbb{Z}_{p-1}$ would come into things.

Answer 1

Let $b \in \mathbb{Z}_p^*$, and let $a \in \mathbb{Z}$ such that $\gcd(a, p - 1) = 1$.

Then by Euler's theorem we have $a^{\varphi(p-1)} \equiv 1 \pmod {p-1}$.

Then $\left(b^{a^{\varphi(p-1)-1}}\right)^a = b^{a^{\varphi(p-1)}} \equiv b \pmod p$ so $b$ is a perfect $a$-th power.