This question originates from Pinter's Abstract Algebra Exercise I5 of Chapter 23.
Suppose $m$ has a primitive root, and let $n$ be relatively prime to $\phi(m)$. (Suppose $n > 0$.) Prove that if $a$ is relatively prime to $m$, then $x^n\equiv a\,(\operatorname{mod}m)$ has a solution.
Attempt:
Given $a$ is relatively prime to $m$ and $m$ has a primitive root, $a$ is a member of some cyclic group $V_m$ of order $\phi(m)$.
Given $n$ is relatively prime to $\phi(m)$, $n$ is invertible in $\mathbb{Z}_{\phi(m)}^*$. Let $t$ be the inverse of $n$ in $\mathbb{Z}_{\phi(m)}^*$, so $nt\equiv 1\,(\operatorname{mod}\phi(m))$. Let $nt=s\cdot\phi(m) + 1$ for some integer $s$.
Suppose $x^n \equiv a\,(\operatorname{mod}m)$. Then \begin{align*} (x^n)^t &\equiv a^t\,(\operatorname{mod}m) \\ x^{s\cdot\phi(m)+1} &\equiv a^t\,(\operatorname{mod}m) \\ x &\equiv a^t\,(\operatorname{mod}m) & x^{\phi(m)}\equiv 1\,(\operatorname{mod}m) \text{ by Euler's theorem} \end{align*} As $V_m$ is cyclic, $a\in V_m$ implies $a^t\in V_m$. This shows a solution exists for $x^n\equiv a\,(\operatorname{mod}m)$.
Questions:
Does this look reasonable?
It seems even if $a$ is not relatively prime to $m$, $x^n\equiv a\,(\operatorname{mod}m)$ would still have a solution of $a^t$ where t is the inverse of $n$ in $\mathbb{Z}_{\phi(m)}^*$. Correct?