In fact, more is true: for any local field $K$, any degree $n$ field extension $L$ of $K$ and any absolute value $|\cdot|$ on $K$, $|N_K^L(\cdot)|^{1/n}$ is the unique absolute value on $L$
which extends $|\cdot|$. In particular, it is a norm on $L$.
In this post I intend to give a proof of this fact which
does not rely on properties of $K$ other than local compactness.

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Let me give some preliminaries.
We regard here a field $F$ and multiplicative function $|\cdot|:F\to [0,\infty)$,
that is a function satisfying $|0|=0$, $|1|=1$ and $|xy|=|x||y|$ for every $x,y\in F$.
For $C\geq 1$ we say that $|\cdot |$ is a
$C$-absolute value if for every $x,y\in F$, $|x+y|\leq C(|x|+|y|)$
and if $C=1$ we simply say that $|\cdot|$ is an absolute value.
The following is well known.

**Lemma 1:** A $2$-absolute value is an absolute value.

It is an easy exercise to check that if $|\cdot|$ is a $C$-absolute value and $\alpha\in(0,1]$ then $|\cdot |^\alpha$ is a $C^\alpha$-absolute value. However, this does not work in general for $\alpha>1$. To remify this, we study a more homogenous condition.
We say that $|\cdot |$ is a $C$-ultra absolute value if for every $x,y\in L$, $|x+y|\leq C\max\{|x|,|y|\}$ and if $C=1$ we say that $|\cdot|$ is an ultra absolute value.
Now we indeed have that if $|\cdot |$ is a $C$-ultra absolute value then for every $\alpha>0$,
$|\cdot |^\alpha$ is a $C^\alpha$-ultra absolute value.
The two definitions relate trivially: a $C$-ultra absolute value is a $C$- absolute while a $C$-absolute value is a $2C$-ultra absolute value.
In particular, every absolute value is a $2$-ultra absolute value.
The following, however, is less trivial.

**Lemma 2:** An absolute value $|\cdot|$ is a $\max\{1,|2|\}$-ultra absolute value.

**Corollary A:** A $C$-ultra absolute value $|\cdot|$ is a $\max\{1,|2|\}$-ultra absolute value.

*Proof:* Set $\alpha=\log_C 2$ and consider the $2$-ultra absolute value $|\cdot |^\alpha$.
It is trivially a $2$-absolute value, thus an actual absolute value by Lemma 1. By Lemma 2 it is a $\max\{1,|2|^\alpha\}$-ultra absolute value.
Taking now the $1/\alpha$-power, we get that $|\cdot|$ is indeed a $\max\{1,|2|\}$-ultra absolute value.


**Corollary B:** A $C$-absolute value $|\cdot|$ is an absolute value iff $|2|\leq 2$. 

*Proof:* If $|\cdot|$ is an absolute value then clearly $|2|=|1+1|\leq |1|+|1|=2$. Assume $|\cdot|$ is a $C$-absolute value and $|2|\leq 2$.
Then $|\cdot|$ is a $2C$-ultra absolute value, thus by Corollary A, it is a $\max\{1,|2|\}$-ultra absolute value,
hence a $2$-ultra absolute value, as $\max\{1,|2|\}\leq 2$. In particular, 
$|\cdot|$ is a $2$-absolute value, thus it is an actual absolute value by Lemma 1.

**Corollary C:** A $C$-absolute value on $F$ which restricts to an absolute value on a subfield is an absolute value on $F$.

*Proof:* This follows from Corollary B, as 2 belongs to the subfield.

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We are now back to the original setting, where $L$ is a finite filed extension of the local field $K$ and $|\cdot|$ is an absolute value on $K$.
We treat $L$ as a locally compact space by identifying it with $K^n$,
noting that the topology is independent of the choice made.
Recall that a proper map is a continuous map for which preimages of precompact sets are precompact. Equivalently, these are maps which are continuous at infinity.

**Lemma 3:** The map $N=N_K^L:L\to K$ is proper.

*Proof:* This follows easily from the continuity of $N$ at 0 and the fact that $N(x^{-1})=N(x)^{-1}$.

**Theorem:** The map $|N(\cdot)|^{1/n}:L\to [0,\infty)$ is an absolute value.

*Proof:* 
The unit ball $B\subset K$ is compact, hence so is $N^{-1}(B)\subset L$
and its shift $1+N^{-1}(B)$. It follows that the image in $[0,\infty)$ under $|N(\cdot)|$ of $1+N^{-1}(B)$ is bounded by some $C$, thus for $z\in L$,
$$ |N(z)|\leq 1 \Rightarrow |N(z)+1|\leq C. $$
It follows that $|N(\cdot)|$ is a $C$-ultra absolute value.
Indeed, for $x,y\in L$, assuming wlog $|x|\leq |y|$ and setting $z=xy^{-1}$ we have 
$$ |N(x+y)|=|N(y)||N(z)+1|\leq C|N(y)||N(z)+1|= C(|N(x)|+|N(y)|).$$
It follows that $|N(\cdot)|^{1/n}$ is a $C^{1/n}$-ultra absolute value,
thus by Corollary C, it is an actual absolute value.