Here is the bare outline of a proof that the absolute Galois group of a finite extension $K$ of $\mathbf{Q}_p$ is finitely generated. Let $\Omega$ be an algebraic closure of $K$, and denote groups such as $\operatorname{Gal}(\Omega|K)$ by $\Gamma(\Omega|K)$.
The profinite group $\Gamma(\Omega|K)$ has the closed normal subgroup $\Gamma(\Omega|K')$ (the ramification group --- a pro-$p$-group), where $K'$ is the maximal tamely ramified extension of $K$ in $\Omega$, giving rise to the short exact sequence $$ 1\to\Gamma(\Omega|K')\to\Gamma(\Omega|K)\to\Gamma(K'|K)\to1. $$ It is well known that the profinite group $\Gamma(K'|K)$ admits the presentation $$ \langle\sigma,\tau\mid\sigma\tau\sigma^{-1}=\tau^q\rangle, $$ where $q=p^f$ is the cardinality of the residue field $k$ of $K$ (see for example a Hasse's book or a paper by Iwasawa in the Transactions). In view of this, it is sufficient to show that the pro-$p$-group $\Gamma(\Omega|K')$ is finitely generated as a closed normal subgroup of $\Gamma(\Omega|K)$.
This would follow if its abelianisation $\Delta(\Omega|K')$ were finitely generated as a $\mathbf{Z}_p[[\Gamma(K'|K)]]$-module, which would in turn follow if $\Delta(\Omega|K')/\Delta(\Omega|K')^p$ were finitely generated as an $\mathbf{F}_p[[\Gamma(K'|K)]]$-module. But such is indeed the case because the $\mathbf{F}_p[\Gamma(K'|K)]$-module $K'^\times/K'^{\times p}$ is finitely generated, completing the proof.