 Solution

Put $\zeta ={\zeta }_{p^n}$, with $n\ge 1$. The shifted cyclotomic polynomial ${\Phi }_{p^n}(1+X)$ is Eisenstein at $p$: its constant term is $p$, and modulo $p$ it is $X^{p^{n-1}(p-1)}$. Therefore $L/{\mathbb{Q}}_p$ is totally ramified of degree $p^{n-1}(p-1)$ and $\pi =\zeta -1$ is a uniformiser.

The Galois group is $(\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }$, with ${\sigma }_a(\zeta )={\zeta }^a$. For $a\ne 1$, let $r=v_p(a-1)$, $0\le r<n$. Then ${\zeta }^{a-1}$ is a primitive $p^{n-r}$th root. Its difference from one is a uniformiser in the corresponding smaller cyclotomic field, and the relative ramification index is $p^r$. HenceThe uniformizer criterion for lower ramification groups now determines every group. Write $U_j=\{a\in (\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }:a\equiv 1(\mathrm{mod}{p}^j)\}$, with $U_n=\{1\}$. Then the lower ramification filtration of a prime-power cyclotomic extension isFor $n=1$ the middle range is empty and the extension is tame. The formula also includes $p=2$; then $U_1$ is already the whole group, so the first possible drop is later than in the odd-prime case. For $p=2,n=1$, the entire extension is trivial.