= Solution
Let $G=\operatorname{Gal}(L/K)$. The lower <ramification groups> are $G_{-1}=G$ and, for $i\geq0$,
$$
G_i=\{\sigma\in G:v_L(\sigma(x)-x)\geq i+1
\text{ for every }x\in\mathcal O_L\}.
$$
If $L/K$ is totally ramified and $\pi_L$ is a <uniformizer>, then $\mathcal O_L=\mathcal O_K[\pi_L]$. Factoring $F(\sigma\pi_L)-F(\pi_L)$ by $\sigma\pi_L-\pi_L$ for $F\in\mathcal O_K[X]$ proves the <uniformizer criterion for lower ramification groups>
$$
G_i=\{\sigma\in G:v_L(\sigma(\pi_L)-\pi_L)\geq i+1\}.
$$
For $\sigma\in G_0$, define
$$
\theta(\sigma)=\overline{\frac{\sigma(\pi_L)}{\pi_L}}\in k_L^\times.
$$
The <inertia group> $G_0$ acts trivially on $k_L$, so $\theta(\sigma\tau)=\theta(\sigma)\theta(\tau)$. Its kernel consists exactly of those $\sigma$ for which $\sigma(\pi_L)/\pi_L\equiv1$ modulo the maximal ideal, namely $G_1$. The <first isomorphism theorem> therefore gives an injection
$$
G_0/G_1\hookrightarrow k_L^\times.
$$
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