 Solution

Put $\alpha =\sqrt[p]{p}$. The polynomial $X^p-p$ is Eisenstein, so ${\mathbb{Q}}_p(\alpha )/{\mathbb{Q}}_p$ is totally ramified of degree $p$. The cyclotomic extension ${\mathbb{Q}}_p({\zeta }_p)/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$. Their coprime degrees make their intersection trivial, sohas degree $p(p-1)$ and is totally ramified. It is the splitting field of $X^p-p$, hence Galois.

Normalize $v_K$ by $v_K(K^{\times })=\mathbb{Z}$. Thensois a uniformizer. Write an automorphism aswhere $a\in {\mathbb{F}}_p^{\times }$ and $b\in {\mathbb{F}}_p$. Sincethe uniformizer criterion for lower ramification groups gives valuation one for $\sigma (\varpi )-\varpi$ when $a\ne 1$, and valuation $p+1$ when $a=1$, $b\ne 0$. Thereforeand $G_i=1$ for $i\ge p+1$. These are the ramification groups of the splitting field of Xp minus p over the p-adic numbers.