Let $\alpha$ be algebraic over $F$. Suppose the degree of $\alpha$ over $F$ is odd. Show that $F(\alpha)=F(\alpha^2)$?

Question

I am trying to solve the following exercise:

Let $\alpha$ be algebraic over $F$. Suppose the degree of $\alpha$ over $F$ is odd. Show that $F(\alpha)=F(\alpha^2)$

It's not clear to me how to solve this. I've been thinking and I guess that $F(\alpha^2)\subset F(\alpha)$ but I don't know about the other direction. Can you help me?

Answer 1

$[F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F]$ since $[F(\alpha):F]$ is odd, we deduce that $[F(\alpha):F(\alpha^2)]$ is odd and is $1$ since the fact $\alpha$ is a solution of $X^2-\alpha^2$ implies that $[F(\alpha):F(\alpha^2)]\leq 2$.