[proofplan] We reduce the claim to two sub-statements: (1) $\mathcal{O}_K$ is finite over $W(\mathbb{F}_q)$, which follows from the structure theorem for complete DVRs (Finite Generation over Witt Vectors), and (2) $W(\mathbb{F}_q)$ is finite over $\mathbb{Z}_p = W(\mathbb{F}_p)$. For the second, we lift an $\mathbb{F}_p$-basis of $\mathbb{F}_q$ to $W(\mathbb{F}_q)$ via the Teichmuller map and show these lifts form a $\mathbb{Z}_p$-basis by the density-plus-completeness argument. Composing the two finite extensions gives $\mathcal{O}_K$ finite over $\mathbb{Z}_p$, from which $K$ is finite over $\mathbb{Q}_p$. [/proofplan] [step:Identify the Witt vector subring $W(\mathbb{F}_q) \hookrightarrow \mathcal{O}_K$ and apply the structure theorem] Let $K$ be a local field of mixed characteristic with residue field $k_K = \mathbb{F}_q$, where $q = p^f$ for some prime $p$ and integer $f \geq 1$. Let $\pi$ be a uniformizer of $\mathcal{O}_K$ and $e = v_K(p)$ the absolute ramification index. By the [Finite Generation over Witt Vectors](/theorems/???), $\mathcal{O}_K$ is a finite free $W(\mathbb{F}_q)$-module of rank $e$: \begin{align*} \mathcal{O}_K = \bigoplus_{i=0}^{e-1} \pi^i W(\mathbb{F}_q). \end{align*} This applies because $\mathcal{O}_K$ is a complete DVR of mixed characteristic $p$ with perfect residue field $\mathbb{F}_q$ and uniformizer $\pi$. [guided] The first ingredient is to identify $W(\mathbb{F}_q)$ inside $\mathcal{O}_K$. Since $\mathbb{F}_q$ is a perfect field of characteristic $p$, the [Witt Vectors](/theorems/???) construction provides a unique strict $p$-ring $W(\mathbb{F}_q)$ with $W(\mathbb{F}_q)/pW(\mathbb{F}_q) \cong \mathbb{F}_q$. The residue field of $\mathcal{O}_K$ is also $\mathbb{F}_q$, and $\mathcal{O}_K$ is itself a strict $p$-ring (it is $p$-torsion free as an integral domain of characteristic $0$, $p$-adically complete, and has perfect residue field $\mathbb{F}_q$). By the [Lifting Homomorphisms Between Strict $p$-Rings](/theorems/???), the identity map $\mathbb{F}_q \to \mathbb{F}_q$ lifts uniquely to a ring homomorphism $W(\mathbb{F}_q) \to \mathcal{O}_K$. This map is injective because its kernel is a prime ideal of $W(\mathbb{F}_q)$ (as $\mathcal{O}_K$ is an integral domain) and cannot be $pW(\mathbb{F}_q)$ (since $p \neq 0$ in $\mathcal{O}_K$), so the kernel is $\{0\}$. With the embedding $W(\mathbb{F}_q) \hookrightarrow \mathcal{O}_K$ established, the [Finite Generation over Witt Vectors](/theorems/???) theorem applies: $\mathcal{O}_K$ is finite free of rank $e$ over $W(\mathbb{F}_q)$, with basis $1, \pi, \ldots, \pi^{e-1}$. [/guided] [/step] [step:Show $W(\mathbb{F}_q)$ is finite free of rank $f$ over $\mathbb{Z}_p = W(\mathbb{F}_p)$] The inclusion $\mathbb{F}_p \hookrightarrow \mathbb{F}_q$ induces, by the functoriality of [Witt Vectors](/theorems/???), an injective ring homomorphism $\mathbb{Z}_p = W(\mathbb{F}_p) \hookrightarrow W(\mathbb{F}_q)$. We claim $W(\mathbb{F}_q)$ is a free $\mathbb{Z}_p$-module of rank $f = [\mathbb{F}_q : \mathbb{F}_p]$. Choose elements $\bar{x}_1, \ldots, \bar{x}_f \in \mathbb{F}_q$ forming an $\mathbb{F}_p$-basis of $\mathbb{F}_q$, and let $x_i = [\bar{x}_i] \in W(\mathbb{F}_q)$ be their Teichmuller lifts. Set $M = \bigoplus_{i=1}^f x_i \mathbb{Z}_p \subset W(\mathbb{F}_q)$. We show $M = W(\mathbb{F}_q)$ by proving $M$ is dense and closed. For density: every element of $W(\mathbb{F}_q)$ has a Witt expansion $\sum_{n=0}^\infty [a_n] p^n$ with $a_n \in \mathbb{F}_q$. Each $[a_n]$ can be written as $\sum_{i=1}^f c_{n,i} [\bar{x}_i]$ modulo $p W(\mathbb{F}_q)$, where $c_{n,i} \in \mathbb{Z}_p$ lift the $\mathbb{F}_p$-coordinates of $a_n$. Iterating this decomposition shows $W(\mathbb{F}_q) = M + p^m W(\mathbb{F}_q)$ for all $m \geq 1$, so $M$ is dense. Since $M$ is a finitely generated module over the $p$-adically complete ring $\mathbb{Z}_p$, $M$ is $p$-adically complete and hence closed. Therefore $M = W(\mathbb{F}_q)$. [guided] We need to show that $W(\mathbb{F}_q)$, which is a priori an abstract ring, is actually a finite free module over $\mathbb{Z}_p$. The idea is to use the residue field structure: $\mathbb{F}_q$ is an $f$-dimensional $\mathbb{F}_p$-vector space, and we lift this structure to the Witt vectors. Pick an $\mathbb{F}_p$-basis $\bar{x}_1, \ldots, \bar{x}_f$ of $\mathbb{F}_q$ and let $x_i = [\bar{x}_i] \in W(\mathbb{F}_q)$ be their Teichmuller representatives. Define $M = \bigoplus_{i=1}^f x_i \mathbb{Z}_p$. We claim $M = W(\mathbb{F}_q)$. **Density.** Take any $a \in W(\mathbb{F}_q)$. By the [Witt Expansion](/theorems/???), $a = \sum_{n=0}^\infty [a_n] p^n$ with $a_n \in \mathbb{F}_q$. For each $n$, write $a_n = \sum_{i=1}^f c_{n,i} \bar{x}_i$ with $c_{n,i} \in \mathbb{F}_p$. The Teichmuller map is multiplicative, so $[a_n] \equiv \sum_{i=1}^f \tilde{c}_{n,i} x_i \pmod{p}$, where $\tilde{c}_{n,i} \in \mathbb{Z}_p$ lifts $c_{n,i}$. (Strictly, this works modulo $p$ because $[a_n] - \sum \tilde{c}_{n,i} [\bar{x}_i]$ reduces to $0$ in $\mathbb{F}_q$.) Thus $a \equiv \sum_{i=1}^f (\sum_n \tilde{c}_{n,i} p^n) x_i \pmod{p^m W(\mathbb{F}_q)}$ for any $m$, giving $W(\mathbb{F}_q) = M + p^m W(\mathbb{F}_q)$ for all $m \geq 1$. **Closedness.** The module $M = \bigoplus_{i=1}^f x_i \mathbb{Z}_p$ is isomorphic to $\mathbb{Z}_p^f$ as a $\mathbb{Z}_p$-module. Since $\mathbb{Z}_p$ is $p$-adically complete, so is $\mathbb{Z}_p^f$ (completeness passes to finite direct sums), and $M$ is $p$-adically complete. A complete submodule of a Hausdorff topological module is closed. Since $M$ is both dense and closed in $W(\mathbb{F}_q)$, we conclude $M = W(\mathbb{F}_q)$. **Linear independence.** The $x_i$ are $\mathbb{Z}_p$-linearly independent: if $\sum_{i=1}^f c_i x_i = 0$ with $c_i \in \mathbb{Z}_p$, reducing modulo $p$ gives $\sum_{i=1}^f \bar{c}_i \bar{x}_i = 0$ in $\mathbb{F}_q$. Since the $\bar{x}_i$ are $\mathbb{F}_p$-linearly independent, $\bar{c}_i = 0$ for all $i$, so $c_i \in p\mathbb{Z}_p$. Dividing by $p$ and repeating, $c_i \in p^n \mathbb{Z}_p$ for all $n$, forcing $c_i = 0$. So $W(\mathbb{F}_q) = \bigoplus_{i=1}^f x_i \mathbb{Z}_p$ is free of rank $f$. [/guided] [/step] [step:Conclude $K$ is a finite extension of $\mathbb{Q}_p$] Combining the two steps: $\mathcal{O}_K$ is free of rank $e$ over $W(\mathbb{F}_q)$, and $W(\mathbb{F}_q)$ is free of rank $f$ over $\mathbb{Z}_p$. By transitivity of freeness, $\mathcal{O}_K$ is a free $\mathbb{Z}_p$-module of rank $ef$. Since $K = \mathcal{O}_K[p^{-1}]$ and $\mathbb{Q}_p = \mathbb{Z}_p[p^{-1}]$, the extension $K/\mathbb{Q}_p$ satisfies \begin{align*} [K : \mathbb{Q}_p] = \operatorname{rank}_{\mathbb{Z}_p}(\mathcal{O}_K) = ef < \infty. \end{align*} Therefore $K$ is a finite extension of $\mathbb{Q}_p$. [/step]