let $k$ be a positive divisor of $q-1$ and $a\in F_q$ be such that the equation $x^k=a$ has no solution in $F_q$. Prove that the same equation has a solution in $F_{q^m}$ if $m$ is divisible by $k$, and that the converse holds for a prime number $k$.
Try using the hypotheses to build the solution, for example: as $ k | q-1 $ so in $ F_q ^ * $ there exists $ \beta $ such that $ \beta ^ k = 1 $ ... but the truth is that I do not know what to do. Help me please.