The equation has a solution iff $(k,q-1)=1$

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Let $p^n$, where $p$ is a prime, and $k \in \mathbb{N}$. Let $a$ the generator of the cyclic group $\mathbb{F}_p^{\star}$, where $\mathbb{F}_q$ the finite field with $q=p^n$ elements. Show that the equaction $x^k-a=0$ has a root in $\mathbb{F}_q$ if and only if $(k,q-1)=1$.

I have done the following:

$$(k, q-1)=1\Leftrightarrow \lambda k+\mu (q-1)=1\Leftrightarrow b^{\lambda k}\cdot (b^{q-1})^{\mu}=b\Leftrightarrow b^{\lambda k}=b\Leftrightarrow (b^{\lambda })^k-b=0$$

Is this correct??