I'm reading a proof of Kronecker-Weber theorem, but I don't know a step in the proof of the below proposition:
Prop. Let p be a prime number, $K/\mathbb Q_p$ be a cyclic extension of degree $l^r$ for some prime $l\neq p$, Then $K\subset \mathbb Q_p(\zeta_m) $ for some $m\in \mathbb Z.$
Proof.(I only quote the part related to my question.)
Let F be the maximal unramified extension of $\mathbb Q_p$ in $K$; then $F$ is cyclotomic, by (a Corollary in the text manual), so $F=\mathbb Q_p(\zeta_n)$. The extension $K/F$ is totally ramified, and it must be tamely ramified, since the ramification index is necessarily a power of $l$ and therefore not divisible by $p$. By (a Theorem in the text manual), we have $K=F(\pi^{1/e})$ for some uniformizer $\pi$ of the discrete valuation ring $O_F$, with $e = [K:F]$. We may assume that $\pi = -pu$ for some $u\in O^{\times}_F$, since $F/\mathbb Q_p$ is unramified.
The extension $F(u^{1/e})/F$ is unramified, since $p\nmid e$ and $u$ is a unit. (the discriminant of $x^e-u$ is not divisible by $p$.)
Why is the second highlighted statement true?
I know a theorem that ramified primes are those divide the discriminant ideal, but $x^e-u$ seems to be required to be irreducible to apply anything I know, but I'm not sure if it's irreduclble or not.
The discriminant of the polynomial $x^e-u$ is $\pm u^{e-1} e^e$, which is a unit, so the extension $F(u^{1/e})/F$ is unramified. This still works if $x^e-u$ is not irreducible, because its field discriminant must be a divisor of $\pm u^{e-1} e^e$.
Another approach to see $F(u^{1/e})/F$ is unramified is to check $x^e-u=0$ has distinct roots in the residual field of $F$, this is evident because $p\nmid e$.