Let me first add that Herbrand wasn't the first to publish his result; it was obtained (but with a less clear exposition) by Pollaczek (*Über die irregulären Kreiskörper der $\ell$-ten und $\ell^2$-ten Einheitswurzeln*, Math. Z. 21 (1924), 1--38).

Next the claim that the class field is generated by a unit is true if $p$ does not divide the class number of the real subfield, that is, if Vandiver's conjecture holds for the prime $p$.

**Proof.** (Takagi)
Let $K = {\mathbb Q}(\zeta_p)$, and assume that the class number of
its maximal real subfield $K^+$ is not divisible by $p$. Then any
unramified cyclic extension $L/K$ of degree $p$ can be written in
the form $L = K(\sqrt[p]{u})$ for some unit $u$ in $O_K^\times$.

In fact, we have $L = K(\sqrt[p]{\alpha})$ for some element
$\alpha \in O_K$. By a result of Madden and Velez, $L/K^+$ is normal
(this can easily be seen directly). If it were abelian, the subextension
$F/K^+$ of degree $p$ inside $L/K^+$ would be an unramified cyclic
extension of $K^+$, which contradicts our assumption that its class
number $h^+$ is not divisible by $p$.

Thus $L/K^+$ is dihedral. Kummer theory demands that
$\alpha /\alpha' = \beta^p$ for some $\beta \in K^+$, where

$\alpha'$ denotes the complex conjugate of $\alpha$.

Since $L/K$ is unramified, we must have $(\alpha) = {\mathfrak A}^p$.
Thus $(\alpha \alpha') = {\mathfrak a}^p$, and since $p$ does not
divide $h^+$, we must have ${\mathfrak a} = (\gamma)$, hence
$\alpha \alpha' = u\gamma^p$ for some real unit $u$.

Putting everything together we get $\alpha^2 = u(\beta\gamma)^p$,
which implies $L = K(\sqrt[p]{u})$.

If $p$ divides the plus class number $h^+$, I cannot exclude the possibility that the Kummer generator is an element that is a $p$-th ideal power, and I cannot see how this should follow from Kummer theory, with or without Herbrand-Ribet.

If $p$ satisfies the Vandiver conjecture, the unit in question can be given explicitly, and was given explicitly already by Kummer for $p = 37$ and by Herbrand for general irregular primes satisfying Vandiver: let $g$ denote a primitive root modulo $p$, and let $\sigma_a: \zeta \to \zeta^a$. Then
$$ u = \eta_\nu = \prod_{a=1}^{p-1} \bigg(\zeta^\frac{1-g}{2}\
\frac{1-\zeta^g}{1-\zeta}\bigg)^{a^\nu \sigma_a^{-1}}, $$
where $\nu$ is determined by $p \mid B_{p-\nu}$.

Here is a survey on class field towers based on my (unpublished) thesis on the explicit construction of Hilbert class fields that I have not really updated for quite some time. Section 2.6 contains the answer to your question for primes satisfying Vandiver.