About primes and cyclotomic extensions

Note for $n\geq 3,\, \Bbb{Q}[\zeta_n]$ is complex and for any $n$ is normal with abelian galois group. Suppose $\sqrt[p]{p} \in \Bbb{Q}[\zeta_n]$. Since $\sqrt[p]{p}$ is real, it is contained in the fixed field of complex conjugation, call it $K$. As $Gal(\Bbb{Q}[\zeta_n])$ is abelian, $K$ is galois hence must be normal. But if $p\geq 3$, $K$ doesn't contain the roots of $x^p-p$ conjugate to $\sqrt[p]{p}$, namely $\zeta_p\sqrt[p]{p},\, \zeta_p^2\sqrt[p]{p},\dots$ since the roots are complex, so it can't be normal. Hence $\sqrt[p]{p} \not \in \Bbb{Q}[\zeta_n]$ for any $p\geq 3$

We use a lemma:

Let $p$ be a prime number, $k\in \mathbb{Q}$, if $x^p-k$ has no rational root, then $x^p-k$ is irreducible over $\mathbb{Q}[x]$

Assume $\sqrt[p]{k} \in \mathbb{Q}(\zeta_n)$. Since the extension $\mathbb{Q}(\zeta_n)$ is normal over $\mathbb{Q}$ and $\sqrt[p]{k}$ is a root of the polynomial $x^p-k$, the lemma says the polynomial $x^p-k$ splits completely in $\mathbb{Q}(\zeta_n)$, hence $\zeta_p \in \mathbb{Q}(\zeta_n)$. Consider the chain of extensions:

$$ F:=\mathbb{Q}\quad \subset \quad L:=\mathbb{Q}(\zeta_p, \sqrt[p]{k}) \quad \subset \quad K:=\mathbb{Q}(\zeta_n)$$

Both extensions $K/F$ and $L/F$ are Galois, the Galois group for $K/F$ is abelian of order $\varphi(n)$, while the Galois group for $L/F$ has order $p(p-1)$, it is a group which is not abelian when $p\geq 3$, (more specifically, it is the general affine group over $\mathbb{F}_p$), a contradiction, hence $\sqrt[p]{k} \notin \mathbb{Q}(\zeta_n)$.