 Solution

If $K$ is complete and $L/K$ is finite, the unique extension of an absolute value to a finite extension isIt restricts to the given absolute value because $N_{L/K}(a)=a^{[L:K]}$ for $a\in K$. The usual construction through multiplication by $x$ on the finite-dimensional $K$-vector space $L$, together with completeness, proves that this is the only extending absolute value.

For $\sigma \in \mathrm{Gal}(L/K)$, the map $x\mapsto |\sigma (x){|}_L$ is another absolute value extending $|\cdot |$ on $K$. Uniqueness therefore gives