Let $p$ be a prime. Let $n$ be a positive integer dividing $p-1$. Suppose that $a^{(p-1)/n}=1$ in the finite field $\mathbb F_p$ with exactly $p$ elements. Then does there exist $b$ with $a=b^n$?
Of course if there exists such an element $b$ then $a^{(p-1)/n}=1$ by Fermat little theorem.
The multiplicative group of $\Bbb F_p$ is cylic (this is not obvious, but it is a well-known result). Say that $g$ is a generator. Then $a=g^k$ for some $k$. Now $a^{(p-1)/n}=1$ implies $$g^{\frac kn(p-1)}=1$$ but since $g$ is a generator, $\frac kn(p-1)$ is a multiple of $p-1$, that is, $n$ divides $k$. Or $k=dn$.
Finally, $a=g^k=g^{dn}=(g^d)^n$. Put $b=g^d$ and you are done.