Proof of Kummer's Lemma in S. Langs 'Cyclotomic fields'

Question

I was going through the proof of Kummer's Lemma (stated below) as done in Serge Langs Cyclotomic fields on page 312.

Now the author states that by class field theory it suffices to show that $\mathbf{Q}(\xi_p, u^{1 / p})$ for a unit $u$ and a primitive $p$-th root of unity $\xi_p$ is unramified over $\mathbf{Q}(\xi_p)$. I do not see how this happens (or: this seems like magic to me) - could anyone explain this in a bit more detail?

Thanks for any help!

Kummer's Lemma: Let $p$ be a regular prime (i.e. an odd prime which does not divide the class number of the $p$-th cyclotomic field) and $\xi$ a primitive $p$-th root of unity; if a unit $u \in \mathbf{Q}(\xi)$ is congruent to an integer modulo $p$, then $u$ is a $p$-th power in $\mathbf{Q}(\xi)$.

Answer 1

Since $K = \mathbf{Q}(\zeta_p)$ contains the $p$th roots of unity, $K(u^{1/p})$ is the splitting field of $x^p - u \in K[x]$. But $x^p - u$ is either irreducible or splits into linear factors depending on whether $u^{1/p} \in K$, hence $K(u^{1/p})/K$ is Galois of degree dividing $p$. Thus the extension is necessarily abelian, so if it's unramified then it sits inside the Hilbert class field of $K$. But, by class field theory, the Hilbert class field of $K$ has degree over $K$ equal to $\# \mathrm{Cl}(K)$. So if $p$ doesn't divide $\# \mathrm{Cl}(K)$ then this would force $K(u^{1/p}) = K$.