[proofplan]
We show $N(L_n/K) = \langle \pi \rangle \times U_K^{(n)}$ by proving two containments. For $\pi \in N(L_n/K)$, we use the norm computation $N_{L_n/K}(-\lambda_n) = \pi$ from the Eisenstein minimal polynomial. For $U_K^{(n)} \subseteq N(L_n/K)$, we use local class field theory: the Artin map identifies $U_K/U_K^{(n)}$ with $\operatorname{Gal}(L_n/K)$, so $U_K^{(n)}$ maps to the routine automorphism and hence lies in the norm group. An index count confirms the containment is an equality.
[/proofplan]
[step:Show $\pi \in N(L_n/K)$]
By the [Galois Group of Lubin--Tate Extensions](/theorems/2394), $\lambda_n \in F(n) \setminus F(n-1)$ is a uniformizer of $L_n$ and its minimal polynomial over $K$ is the Eisenstein polynomial $\phi_n(X) = e^n(X)/e^{n-1}(X)$. The constant term of $\phi_n$ is $\pi$ (this follows from the recursive structure: the product of all roots of $e^n$ over those of $e^{n-1}$ contributes a factor of $\pi$). Therefore
\begin{align*}
N_{L_n/K}(-\lambda_n) = (-1)^{[L_n:K]} \cdot (\text{constant term of } \phi_n) = \pi,
\end{align*}
where the sign works out because the constant term of an Eisenstein polynomial $X^d + a_{d-1}X^{d-1} + \cdots + a_0$ has $N_{L_n/K}(\alpha) = (-1)^d a_0$ for a root $\alpha$, and here $(-1)^d a_0 = \pi$. Hence $\pi \in N_{L_n/K}(L_n^\times) = N(L_n/K)$.
[/step]
[step:Show $U_K^{(n)} \subseteq N(L_n/K)$]
By the [Galois Group of Lubin--Tate Extensions](/theorems/2394), the Galois group satisfies
\begin{align*}
\operatorname{Gal}(L_n/K) \cong U_K / U_K^{(n)},
\end{align*}
where a unit $u \in U_K$ acts on $F(n)$ by $\lambda \mapsto [u]_F(\lambda)$. Local class field theory identifies $K^\times / N(L_n/K) \cong \operatorname{Gal}(L_n/K)$ via the Artin map, which is compatible with the Lubin--Tate isomorphism: the Artin map sends $u \in U_K$ to $\sigma_{u^{-1}} \in \operatorname{Gal}(L_n/K)$.
An element $u \in U_K$ lies in $N(L_n/K)$ if and only if $\operatorname{Art}_K(u)|_{L_n} = \operatorname{id}_{L_n}$, i.e., $\sigma_{u^{-1}} = \operatorname{id}$, i.e., $u^{-1} \in U_K^{(n)}$, i.e., $u \in U_K^{(n)}$ (since $U_K^{(n)}$ is a group). Therefore
\begin{align*}
U_K \cap N(L_n/K) = U_K^{(n)}.
\end{align*}
[guided]
The Artin reciprocity map provides a canonical isomorphism $K^\times/N(L_n/K) \xrightarrow{\sim} \operatorname{Gal}(L_n/K)$. The Lubin--Tate theory gives a separate isomorphism $U_K/U_K^{(n)} \xrightarrow{\sim} \operatorname{Gal}(L_n/K)$. The key fact (established in the construction of the explicit local Artin map) is that these are compatible: the restriction of $\operatorname{Art}_K$ to $U_K$ sends $u$ to $\sigma_{u^{-1}}$, which acts directly on $L_n$ if and only if $u^{-1} \equiv 1 \pmod{\pi^n \mathcal{O}_K}$, i.e., $u \in U_K^{(n)}$.
Therefore $U_K^{(n)}$ is precisely the kernel of $\operatorname{Art}_K|_{U_K}: U_K \to \operatorname{Gal}(L_n/K)$, which equals $U_K \cap N(L_n/K)$ by the definition of the Artin map.
[/guided]
[/step]
[step:Combine to conclude $N(L_n/K) = \langle \pi \rangle \times U_K^{(n)}$]
From the first step, $\pi \in N(L_n/K)$. Since every element of $K^\times$ can be written as $\pi^m u$ with $m \in \mathbb{Z}$ and $u \in U_K$, we have $\pi^m u \in N(L_n/K)$ if and only if $\pi^m \in N(L_n/K)$ and $u \in N(L_n/K)$. Since $\pi \in N(L_n/K)$, the condition $\pi^m \in N(L_n/K)$ holds for all $m \in \mathbb{Z}$. The condition on $u$ is $u \in U_K^{(n)}$ by the second step.
Therefore $N(L_n/K) \supseteq \langle \pi \rangle \times U_K^{(n)}$.
For the reverse containment, we verify the index. The group $K^\times / (\langle \pi \rangle \times U_K^{(n)}) \cong U_K / U_K^{(n)}$ has order $q^{n-1}(q-1) = [L_n : K]$. By local class field theory, $(K^\times : N(L_n/K)) = [L_n : K]$ for an abelian extension. Since $\langle \pi \rangle \times U_K^{(n)} \subseteq N(L_n/K)$ and both have the same index $[L_n : K]$ in $K^\times$, we conclude
\begin{align*}
N(L_n/K) = \langle \pi \rangle \times U_K^{(n)}.
\end{align*}
[/step]
