Prove that if $[F[u]:F]$ is odd, then $F[u]=F[u^2]$

Question

Let $F$ be a field, prove that if $[F[u]:F]$ is odd, where $[F[u]:F]$ is the degree of the minimum polynomial of $u$ over $F$, then $F[u]=F[u^2]$.

I state that since $u$ is algebraic over $F$, let $f(x)=x^n+a_1x^{n-1}+\dots+a_n\in F[x]$ be the minimum polynomial of $u$ over $F$ (where $n$ is odd), so if $g(x)\in F[x]$, $g(x)=f(x)q(x)+r(x)$ where $\deg r(x)<\deg f(x)=n$ and we can write $g(u)=0\cdot q(u)+r(u)$, hence the elements of $F[u]$ has the form $r(u)=b_0+b_1u+\dots+b_{n-1}u^{n-1}$. At this point of the proof i don't know how to go on. Any ideas?

Answer 1

Hint 1:

If $F\subset F[u^2]\subset F[u]$ (proper inclusion) then $\left [F[u]:F\right ]=\left [F[u]:F[u^2]\right ]\cdot \left [F[u^2]:F\right ]$

Hint 2:

What are the possible values for $\left [F[u]:F[u^2]\right ]$?

Hint 3:

Assuming $\left [F[u]:F\right ]$ is odd, try to reach a contradiction.