= Solution
Put $\zeta=\zeta_{p^n}$, with $n\geq1$. 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 $a\ne1$, let $r=v_p(a-1)$, $0\leq r<n$. 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 $p^r$. Hence
$$
v_L(\sigma_a(\pi)-\pi)=v_L(\zeta^{a-1}-1)=p^r.
$$
The <uniformizer criterion for lower ramification groups> now determines every group. Write $U_j=\{a\in(\mathbb Z/p^n\mathbb Z)^\times:a\equiv1\pmod{p^j}\}$, with $U_n=\{1\}$. Then the <lower ramification filtration of a prime-power cyclotomic extension> is
$$
\boxed{\begin{aligned}
G_{-1}=G_0&=(\mathbb Z/p^n\mathbb Z)^\times,\\
G_i&=U_j&&\text{if }1\leq j\leq n-1,\quad p^{j-1}\leq i\leq p^j-1,\\
G_i&=\{1\}&&\text{if }i\geq p^{n-1}.
\end{aligned}}
$$
For $n=1$ 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.
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