[proofplan]
We derive the formula $e_{L/K} = (\mathcal{O}_K^\times : N_{L/K}(\mathcal{O}_L^\times))$ from local class field theory by analyzing the exact sequence relating $K^\times / N(L/K)$, the valuation map, and the unit norm index. The key steps are: (1) use the valuation to produce a surjection $K^\times / N(L/K) \twoheadrightarrow \mathbb{Z} / f_{L/K} \mathbb{Z}$; (2) identify its kernel as $\mathcal{O}_K^\times / N_{L/K}(\mathcal{O}_L^\times)$; (3) apply local class field theory to compute $(K^\times : N(L/K)) = [L:K]$; and (4) divide by $f_{L/K}$ to isolate the unit norm index.
[/proofplan]
[step:Establish the surjection from $K^\times / N(L/K)$ to $\mathbb{Z} / f_{L/K} \mathbb{Z}$ via the valuation]
Let $w$ be the unique extension of the valuation $v_K$ to $L$, normalized so that $w|_K = v_K$. For any $x \in L^\times$, the norm satisfies
\begin{align*}
v_K(N_{L/K}(x)) = f_{L/K} \cdot v_L(x),
\end{align*}
where $v_L = e_{L/K} \cdot w$ is the normalized valuation on $L$ (with $v_L(L^\times) = \mathbb{Z}$). Since $v_L$ is surjective onto $\mathbb{Z}$, the image $v_K(N(L/K)) = f_{L/K} \mathbb{Z}$.
The valuation $v_K: K^\times \to \mathbb{Z}$ therefore induces a surjection
\begin{align*}
\bar{v}_K: K^\times / N(L/K) \twoheadrightarrow \mathbb{Z} / f_{L/K} \mathbb{Z}.
\end{align*}
[guided]
The formula $v_K(N_{L/K}(x)) = f_{L/K} \cdot v_L(x)$ comes from the standard relation between the valuation and the norm: if $w$ extends $v_K$ with $v_K = w|_K$, then $v_K(N_{L/K}(x)) = [L:K] \cdot w(x)$, and $v_L(x) = e_{L/K} \cdot w(x)$, so $v_K(N_{L/K}(x)) = [L:K] / e_{L/K} \cdot v_L(x) = f_{L/K} \cdot v_L(x)$.
Since $v_L$ surjects onto $\mathbb{Z}$ (by the normalization of $v_L$), the image of $N(L/K)$ under $v_K$ is exactly $f_{L/K} \mathbb{Z}$. This means $v_K$ descends to a well-defined surjection on the quotient $K^\times / N(L/K)$.
[/guided]
[/step]
[step:Identify the kernel of $\bar{v}_K$ as $\mathcal{O}_K^\times / N_{L/K}(\mathcal{O}_L^\times)$]
The kernel of $\bar{v}_K$ consists of those cosets $a \cdot N(L/K) \in K^\times / N(L/K)$ with $v_K(a) \in f_{L/K} \mathbb{Z} = v_K(N(L/K))$. Equivalently, $a = u \cdot n$ for some $u \in \mathcal{O}_K^\times$ (the unit part of $a$) and $n \in N(L/K)$ with $v_K(n) = v_K(a)$. In other words, $a \in \mathcal{O}_K^\times \cdot N(L/K)$.
Therefore
\begin{align*}
\ker(\bar{v}_K) = \mathcal{O}_K^\times \cdot N(L/K) / N(L/K) \cong \mathcal{O}_K^\times / (\mathcal{O}_K^\times \cap N(L/K)) = \mathcal{O}_K^\times / N_{L/K}(\mathcal{O}_L^\times),
\end{align*}
where the last equality uses $\mathcal{O}_K^\times \cap N(L/K) = N_{L/K}(\mathcal{O}_L^\times)$: an element of $\mathcal{O}_K^\times$ in the norm group must be the norm of a unit (since $v_K(N_{L/K}(x)) = 0$ forces $v_L(x) = 0$, so $x \in \mathcal{O}_L^\times$).
The isomorphism is the second isomorphism theorem for groups.
[/step]
[step:Apply local class field theory and conclude]
By local class field theory, since $L/K$ is a finite abelian extension of degree $n = [L:K]$:
\begin{align*}
(K^\times : N(L/K)) = [L:K] = n = e_{L/K} \cdot f_{L/K}.
\end{align*}
From the short exact sequence
\begin{align*}
1 \to \ker(\bar{v}_K) \to K^\times / N(L/K) \xrightarrow{\bar{v}_K} \mathbb{Z} / f_{L/K} \mathbb{Z} \to 0,
\end{align*}
we obtain
\begin{align*}
(K^\times : N(L/K)) = |\ker(\bar{v}_K)| \cdot |\mathbb{Z} / f_{L/K} \mathbb{Z}| = (\mathcal{O}_K^\times : N_{L/K}(\mathcal{O}_L^\times)) \cdot f_{L/K}.
\end{align*}
Substituting $(K^\times : N(L/K)) = e_{L/K} \cdot f_{L/K}$ and dividing both sides by $f_{L/K}$:
\begin{align*}
e_{L/K} = (\mathcal{O}_K^\times : N_{L/K}(\mathcal{O}_L^\times)).
\end{align*}
[/step]
