To prove any real Galois extension of $\mathbb{Q}$ of odd degree must not be radical I'm trying to prove this using the statement: 
Let $K$ be a field containing n distinct roots of unity. An extension of $K$ of degree $n$ is a radical extension generated by an $n$th root of an element of $K$ if and only if it is a Galois extension whose Galois group is a cyclic group of order $n$.
but I can't find the link between radical extension and odd degree of the Galois extension 
 A: Suppose $L$ is a non-trivial real radical Galois extension of $\mathbb Q$ of odd degree. 
(I added the condition "non-trivial" to rule out the case where $L = \mathbb Q$.)
Since $L : \mathbb Q$ is radical of degree $> 1$, there exists a tower of fields
$$ E_0 = \mathbb Q \subsetneq E_1 \subset \dots  \subsetneq  E_{n-1}\subsetneq E_n = L$$
with $n \geq 1$, where, for each $i \in \{1, \dots, n \}$, there exist $\alpha_i \in E_{i}$, $\beta_i \in E_{i-1}$, $k_i \in \mathbb N$ such that $$\alpha_i^{k_i} = \beta_i , \ \ \ E_i = E_{i-1}(\alpha_i).$$
Since $[L : \mathbb Q]$ is odd, $[E_1 : \mathbb Q]$ is also odd, by the tower law. As $[E_1 : \mathbb Q] > 1$ by assumption, this implies that $$[E_1 : \mathbb Q] \geq 3.$$
(The same applies to $[E_i : E_{i-1}]$ for each $i \in \{1, \dots, n \}$, but we won't need this.)
Let $m_1(X) \in \mathbb Q[X]$ be the minimal polynomial for $\alpha_1$ over $\mathbb Q$. Then $m_1(X)$ is a polynomial of degree $\geq 3 $, irreducible over $\mathbb Q$, which divides the polynomial $X^{k_1} - \beta_1$.
Since $L$ is Galois over $\mathbb Q$, and since $L$ contains the root $\alpha_1$ of $m_1(X)$, the polynomial $m_1(X)$ splits completely in $L$. The roots of $m_1(X)$ in $L$ must also be roots of $X^{k_1} - \beta_1$, i.e. each root of $m_1(X)$ is equal to $\alpha_1 $ multiplied by some $k_1$th root of unity. As $m_1(X)$ has at least three distinct roots in $L$, at least one of these roots must be non-real. This contradicts the assumption that $L$ is a real field.
