Let $K, L$ be fields, $K \subseteq L$ and $[K(\alpha) : K] = p$ for a prime number $p ≠ 2$, and some $\alpha \in L \backslash K$ that is algebraic over $L$. I now want to show that $K(\alpha^2) = K(\alpha)$.
It's obvious that $K(\alpha^2) \subseteq K(\alpha)$, since every field containing $K$ and $\alpha$ also contains $\alpha^2$, but I couldn't find a way to approach the other inclusion so far. I guess that at some point, I have to use that the degree is an uneven prime number, but I fail to see how that can come into play.
By the tower law, we have $$p = [K(\alpha) : K] = [K(\alpha) : K(\alpha^2)][K(\alpha^2) : K].$$
Note that $\alpha$ is a root of the polynomial $f(x) = x^2 - \alpha^2$ in $K(\alpha^2)[x]$, so $[K(\alpha) : K(\alpha^2)]$ is either $1$ or $2$. Now use the fact that $p \neq 2$ is prime.