Norm group of $\mathbb{Q}_p(\sqrt[4]{\zeta_{p-1}p})$ Let $p \equiv 3\pmod{4}$ be a prime, and let $\zeta_{p-1}$ be a primitive $(p-1)^\mathrm{st}$ root of unity in the $p$-adic number $\mathbb{Q}_p$. Let $K$ be the extension $\mathbb{Q}_p(\sqrt[4]{\zeta_{p-1}p})$. Formally, we define $K$ to be the quotient ring.
$$
K = \frac{\mathbb{Q}_p[X]}{(X^4 - \zeta_{p-1}p)}.
$$
I would like to compute the norm group
$$
\operatorname{Nm}(K^\times) = \{N_{K/\mathbb{Q}_p}(\alpha) : \alpha \in K\}.
$$
My Attempt
Clearly $\mathbb{Q}_p^{\times 4} \subseteq \operatorname{Nm}(K^\times)$, and we also have
$$
N_{K/\mathbb{Q}_p}(\sqrt[4]{\zeta_{p-1}p}) = - \zeta_{p-1}p,
$$
so it follows that
$$
\{u(-\zeta_{p-1})^kp^k : u \in \mathcal{O}_K^{\times 4}, k \in \mathbb{Z}\} \subseteq \operatorname{Nm}(K^\times).
$$
Now, if $K/\mathbb{Q}_p$ were an abelian extension, then we would be done, because local class field theory tells us that $[\mathbb{Q}_p^\times : \operatorname{Nm}(K^\times)] = 4$, and the subgroup $\{u(-\zeta_{p-1})^kp^k : u \in \mathcal{O}_K^{\times 4}, k \in \mathbb{Z}\}$ has index $4$.
However, for $p\equiv 3\pmod{4}$, the extension is not Galois and its Galois closure has Galois group $D_8$ (based on looking at examples in the LMFDB), which obviously is not abelian. Is there still some similar way to conclude the argument?

The answer is given by Theorem III.3.5 of Milne's Class Field Theory notes, which states that for a finite extension $L/K$ of local fields, we have
$$
N_{L/K}(L^\times) = N_{E/K}(E^\times),
$$
where $E$ is the largest abelian extension of $K$ contained in $L$. Therefore in my concrete case, we would have $E = \mathbb{Q}_p(\sqrt{\zeta_{p-1}p})$, so the norm group we are looking for is
$$
\{u(-\zeta_{p-1}p)^k : u \in \mathbb{Z}_p^{\times 2}, k \in \mathbb{Z}\},
$$
which is of index $2$.
