Simple algebraic field extensions

Question

Let $u$ be algebraic over a field $F$ such that $[F(u): F] = n$, and let $m$ be a natural number such that $(n,m!)=1$. Prove $F(u) = F(u^m)$.

I know that since $u$ is algebraic over $F$, then $u$ satisfies $f(u) = 0$, were deg$f$ =n, and $f(x) \in F[x]$. We have $(n,m!)$=1, which means that $\exists x,y \in \mathbb{Z}$ such that $nx + m!y = 1$, but I'm not sure where to go from here.

Answer 1

Already you have that $F(u^m) \subseteq F(u)$. So all you have to do is show that $[F(u) : F(u^m)] = 1$. Let $v = u^m$, and let $\mu(X)$ be the minimal polynomial of $u$ over the field $F(v)$. Then $F(u) = F(v)(u)$, and $[F(v)(u) : F(v)] = [F(u) : F(v)]$ is the same as the degree of $\mu$.

Let $f(X) = X^m - v$. This is polynomial in $F(v)[X]$. Since $u$ is a root of $f$, we can conclude that $\mu$ divides $f$, and in particular the degree of $\mu$ is $\leq$ that of $f$. So, $deg(\mu)$ divides $m!$, but being equal to $[F(u) : F(v)]$, it also divides $[F(u) : F] = n$. Therefore $deg(\mu)$ is equal to $1$, as required.

Answer 2

$F(u^m)\subseteq F(u)$, and $u$ is a root of $X^m-u^m$. This shows that $[F(u):F(u^m)]\le m$. Furthermore, $[F(u):F(u^m)]\mid [F(u):F]=n$. Can you conclude from here?