Write $e = e_{L/K}$ and $f = f_{L/K}$. Since $k_L/k_K$ is a finite separable extension (finite fields are perfect), the primitive element theorem gives $\bar\alpha \in k_L$ with $k_L = k_K(\bar\alpha)$. Let $\bar{f}(x) \in k_K[x]$ be the minimal polynomial of $\bar\alpha$, and let $f(x) \in \mathcal{O}_K[x]$ be a monic lift of $\bar{f}(x)$ of the same degree. We find $\alpha \in \mathcal{O}_L$ lifting $\bar\alpha$ with $v_L(f(\alpha)) = 1$. Any lift $\beta$ of $\bar\alpha$ satisfies $v_L(f(\beta)) \geq 1$ (since $\bar{f}(\bar\alpha) = 0$). If equality holds, set $\alpha = \beta$. Otherwise, set $\alpha = \beta + \pi_L$. By Taylor expansion, $f(\beta + \pi_L) = f(\beta) + f'(\beta)\pi_L + b\pi_L^2$ for some $b \in \mathcal{O}_L$. Since $\bar{f}$ is separable, $f'(\bar\alpha) \neq 0$, so $v_L(f'(\beta)) = 0$. If $v_L(f(\beta)) \geq 2$ then the leading term is $f'(\beta)\pi_L$, giving $v_L(f(\alpha)) = 1$. In all cases, $f(\alpha)$ is a uniformizer $\pi_L = f(\alpha)$ of $L$. We claim $\{\alpha^i \pi_L^j : 0 \leq i \leq f-1,\, 0 \leq j \leq e-1\}$ is an $\mathcal{O}_K$-basis of $\mathcal{O}_L$. For linear independence: suppose $\sum_{i,j} a_{ij}\alpha^i\pi_L^j = 0$ with $a_{ij} \in K$. Set $s_j = \sum_i a_{ij}\alpha^i$. Since $1, \bar\alpha, \ldots, \bar\alpha^{f-1}$ are $k_K$-linearly independent, the reductions of $1, \alpha, \ldots, \alpha^{f-1}$ are linearly independent over $k_K$, which forces $v_L(s_j) \in e\mathbb{Z}$ for each $j$ with $s_j \neq 0$ (a careful valuation argument using the maximal coefficient). Then $v_L(s_j \pi_L^j) \in j + e\mathbb{Z}$, and no two non-zero terms have the same valuation, so $\sum_j s_j\pi_L^j \neq 0$, a contradiction. For spanning: since $1, \bar\alpha, \ldots, \bar\alpha^{f-1}$ span $k_L$ over $k_K$, we get $\mathcal{O}_L = N + \pi_L\mathcal{O}_L$ where $N = \bigoplus_{i=0}^{f-1}\mathcal{O}_L\alpha^i$. Iterating and using $\pi_K \sim \pi_L^e$ (up to a unit), we find $M = \bigoplus_{i,j}\mathcal{O}_K\alpha^i\pi_L^j$ is dense in $\mathcal{O}_L$. As the closed unit ball in a finite-dimensional $K$-subspace, $M$ is complete, hence $M = \mathcal{O}_L$. Finally, since $\alpha^i\pi_L^j = \alpha^i f(\alpha)^j \in \mathcal{O}_K[\alpha]$, we have $\mathcal{O}_L = \mathcal{O}_K[\alpha]$.