I think that the following statement is correct.
Let $n \geq 3$ and let $p$ be prime. Show that $\sqrt[n]{p}$ is not contained in a cyclotomic extension of $\mathbb{Q}$.
My idea for a proof is that, when $n \geq 3$, the minimal polynomial of $\sqrt[n]{p}$ over $\mathbb{Q}$, $x^n-p$ has complex roots (since they are of the form $\zeta_n\cdot\sqrt[n]{p}$). Since $\sqrt[n]{p}$ is real, we have that $\mathbb{Q}(\sqrt[n]{p})$ contains $\sqrt[n]{p}$ but not the complex roots of its minimal polynomial. This implies that $\mathbb{Q}(\sqrt[n]{p})$ is not normal.
Therefore, $\mathbb{Q}(\sqrt[n]{p})$ cannot be an intermediate extension of any cyclotomic extension, since cyclotomic extensions are abelian (all intermediate extensions are normal). This implies that $\sqrt[n]{p}$ is not contained in any cyclotomic extension of $\mathbb{Q}$.
My question is whether this statement is correct, and if so, if my proof is correct. Moreover, does it hold more generally for cyclotomic extensions over fields with characteristic 0? Is $p$ prime necessary, or does it work for any $p \in \mathbb{Q}$ such that $\sqrt[n]{p}$ is irrational?