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Here is a completely elementary proof of the fact that a finite extension $F$ of $\mathbf{Q}_p$ has only finitely many extensions of any given degree $n>0$. As $F$ has a unique unramified extension of any given degree, we are reduced to the case of totally ramified extensions. Next, it is easily seen that there are only finitely many (totally ramified) extensions of degree prime to $p$ (essentially because $F^\times/F^{\times m}$ is finite for any $m$ prime to $p$), so we are reduced to the case $n=p^r$. Passing to the galoisian closure and using the fact that any $p$-group admits a filtration whose successive quotients are of order $p$, we are reduced to the case $n=p$. Now, degree-$p$ extensions of $F$ become cyclic when translated to $K$, the extension of $F$ obtained by adjoining to $F$ the $(p-1)$-th roots of everything in $F$; this follows from "Galois's last theorem" (MO24081), see for example arXiv:1005.2016v3. Two degree-$p$ extensions of $F$ give rise to the same cyclic extension of $K$ if and only if they are conjugate over $F$, and the number of conjugates of a degree-$p$ extension $E$ is $1$ (if $E$ is cyclic over $F$) or exactly $p$ (otherwise). As $K$ automatically contains a primitive $p$-th root of $1$, cyclic extensions of $K$ correspond to $\mathbf{F}_p$-lines in $K^\times/K^{\times p}$, which is finite. In fact, the compositum of all degree-$p$ extensions of $F$ is the extension of $K$ obtained by adjoining the $p$-th roots of everything in $K$. Done. (It is at this point that the proof breaks down if $F$ had been a finite extension of $\mathbf{F}_p((t))$; such $F$ have infinitely many separable (indeed cyclic) extensions of degree $p$.)