I'm struggling with the following problem.
Let n be a natural number, let $F$ be a field that contains a primitive $p^n$-th root of unity and let $a \in F^{\times}$. Show that if deg$(F(\sqrt[p]{a})/F) >1$, then deg$(F(\sqrt[p^n]{a})/F)=p^n$.
I was able to show that deg$(F(\sqrt[p^n]{a})/F)>1$ (this is because deg$(F(\sqrt[p]{a})/F) >1$) and that $F(\sqrt[p^n]{a})/F$ is Galois ($F(\sqrt[p^n]{a})/F$ is separable because $F$ contains a primitive $p^n$-th root of unity and normal because $X^{p^n} -a$ splits over $F(\sqrt[p^n]{a})/F)$).
I thus tried to show that |Gal($F(\sqrt[p^n]{a})/F)| = p^n$ since |Gal($F(\sqrt[p^n]{a})/F)| =$ deg$(F(\sqrt[p^n]{a})/F)$. But this is where I failed (I tried to show that the map Gal($F(\sqrt[p^n]{a})/F \rightarrow \mu_{p^n}$ is surjective).
If someone could help me with a hint, that would be kind.