= Solution
Put $\alpha=\sqrt[p]{p}$ and $\lambda=\zeta_p-1$. The polynomial $T^p-p$ is <Eisenstein>, so $[\mathbb Q_p(\alpha):\mathbb Q_p]=p$, while the preceding part gives degree $p-1$ for $E$. Both numbers divide $[K:\mathbb Q_p]$ by the <tower law>, and the <compositum> degree is at most their product. Coprimality forces $[K:\mathbb Q_p]=p(p-1)$. Since $K$ contains all the roots $\zeta_p^a\alpha$ of $T^p-p$, it is the <splitting field> of a <separable polynomial> and is <Galois>.
Extend $v_p$ with $v_p(p)=1$. Then $v_p(\alpha)=1/p$, $v_p(\lambda)=1/(p-1)$, and
$$
\boxed{\varpi=\frac{\zeta_p-1}{\sqrt[p]{p}},\qquad v_p(\varpi)=\frac1{p(p-1)}.}
$$
The value group therefore forces the ramification index to be at least $p(p-1)$; it cannot exceed the extension degree. Thus $K/\mathbb Q_p$ is totally ramified and $\varpi$ is a <uniformizer> for the normalized valuation $v_K=p(p-1)v_p$.
Write its automorphisms as $\sigma_{a,b}(\alpha)=\zeta_p^a\alpha$ and $\sigma_{a,b}(\zeta_p)=\zeta_p^b$, with $a\in\mathbb F_p$, $b\in\mathbb F_p^\times$. The degree count makes all $p(p-1)$ pairs occur, giving $G\cong\mathbb F_p\rtimes\mathbb F_p^\times$. Their action on the uniformizer is
$$
\frac{\sigma_{a,b}(\varpi)}{\varpi}=\zeta_p^{-a}\frac{\zeta_p^b-1}{\zeta_p-1}.
$$
If $b\ne1$, the residue of this ratio is $b$, so subtracting one gives a unit and $v_K(\sigma_{a,b}(\varpi)-\varpi)=1$. If $b=1$ and $a\ne0$, the ratio is $\zeta_p^{-a}$, giving displacement valuation $1+v_K(\zeta_p^{-a}-1)=p+1$. The identity has infinite displacement valuation.
By the <uniformizer criterion for lower ramification groups>, $\sigma\in G_i$ precisely when that displacement valuation is at least $i+1$. Hence the <ramification groups of the splitting field of Xp minus p over the p-adic numbers> have sizes
$$
\boxed{|G_i|=\begin{cases}p(p-1),&i=-1,0,\\p,&1\le i\le p,\\1,&i\ge p+1.\end{cases}}
$$
For $1\le i\le p$ the group itself is $\{\sigma_{a,1}:a\in\mathbb F_p\}\cong C_p$. The lower wild ramification break is $p$.
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