[proofplan] We establish the proportionality $w = e_{L/K}^{-1} v_L$ by decomposing an arbitrary element $x \in L^\times$ as $x = u\pi_L^m$ with $u \in \mathcal{O}_L^\times$ and computing $w(x)$ using the multiplicativity of the valuation. The minimum formula follows from the fact that $\mathfrak{m}_L = \pi_L \mathcal{O}_L$ consists of elements with $v_L$-valuation $\geq 1$. [/proofplan] [step:Express $w$ in terms of $v_L$ using the uniformizers] By definition, $e_{L/K} = v_L(\pi_K)$, and by the normalisation of $w$, $w(\pi_K) = 1$. Since $\pi_L$ generates the maximal ideal of $\mathcal{O}_L$ and $\pi_K = u\pi_L^{e_{L/K}}$ for some unit $u \in \mathcal{O}_L^\times$, we have $v_L(\pi_L) = 1$ and \begin{align*} w(\pi_L) = \frac{w(\pi_K)}{e_{L/K}} \cdot \frac{v_L(\pi_K)}{v_L(\pi_K)} = \frac{1}{e_{L/K}}. \end{align*} More directly: $1 = w(\pi_K) = w(u) + e_{L/K} \cdot w(\pi_L)$. Since $u \in \mathcal{O}_L^\times$ has $v_L(u) = 0$, and $w$ is a non-archimedean valuation on $L$ extending $v_K$, we have $w(u) = 0$ (the unit group of $\mathcal{O}_L$ with respect to $w$ consists of elements with $w$-value $0$). Therefore $w(\pi_L) = 1/e_{L/K}$. [/step] [step:Verify the proportionality for all $x \in L^\times$] Let $x \in L^\times$. Write $v_L(x) = m \in \mathbb{Z}$, so $x = u\pi_L^m$ for a unique unit $u \in \mathcal{O}_L^\times$. By the multiplicativity of $w$: \begin{align*} w(x) = w(u) + m \cdot w(\pi_L) = 0 + m \cdot \frac{1}{e_{L/K}} = \frac{v_L(x)}{e_{L/K}}. \end{align*} This establishes $w(x) = e_{L/K}^{-1} v_L(x)$ for all $x \in L^\times$. [/step] [step:Derive the minimum formula for the maximal ideal] The maximal ideal $\mathfrak{m}_L = \pi_L \mathcal{O}_L = \{x \in \mathcal{O}_L : v_L(x) \geq 1\}$. By the proportionality established above, for $x \in \mathfrak{m}_L \setminus \{0\}$: \begin{align*} w(x) = \frac{v_L(x)}{e_{L/K}} \geq \frac{1}{e_{L/K}}, \end{align*} with equality when $v_L(x) = 1$, i.e., when $x$ is an associate of $\pi_L$. In particular, $w(\pi_L) = 1/e_{L/K}$ achieves this minimum. Therefore \begin{align*} \min\{w(x) : x \in \mathfrak{m}_L\} = e_{L/K}^{-1}. \end{align*} [guided] We want to compute $\min\{w(x) : x \in \mathfrak{m}_L \setminus \{0\}\}$. By definition, $\mathfrak{m}_L = \{x \in \mathcal{O}_L : v_L(x) \geq 1\}$. Since $v_L$ takes integer values on $L^\times$ and $v_L(x) \geq 1$ for $x \in \mathfrak{m}_L \setminus \{0\}$, the smallest possible value of $v_L(x)$ on $\mathfrak{m}_L \setminus \{0\}$ is $1$, achieved by any uniformizer $x = u\pi_L$ with $u \in \mathcal{O}_L^\times$. Applying the proportionality $w(x) = v_L(x)/e_{L/K}$ from the previous step, the smallest $w$-value on $\mathfrak{m}_L \setminus \{0\}$ is $1/e_{L/K}$, achieved at $x = \pi_L$. Why does the minimum exist? Because $v_L$ is discrete (takes values in $\mathbb{Z}$), the set $\{v_L(x) : x \in \mathfrak{m}_L \setminus \{0\}\} = \{1, 2, 3, \ldots\}$ has a minimum of $1$. The proportionality then transfers this to a minimum of $1/e_{L/K}$ for $w$. Geometrically, this minimum formula tells us that $w$ is the unique extension of $v_K$ to $L$ (normalised to take integer values on $K^\times$) scaled by $e_{L/K}^{-1}$ on $\mathcal{O}_L^\times$-orbits of $\pi_L$. [/guided] [/step]