= Solution
Let $\pi\in K$ be a <uniformizer>. Because $L/K$ is <unramified>, the same $\pi$ is a <uniformizer> of $L$, and $v_L|_K=v_K$. The conjugates of $z\in L^\times$ all have the same <valuation>, so
$$
v_K(N_{L/K}z)=n\,v_L(z).
$$
This proves that every <field norm> has valuation divisible by $n$. Conversely $N_{L/K}(\pi^r)=\pi^{nr}$ for every integer $r$, so it remains to prove <surjectivity of unit norms in unramified extensions>.
Write $|k_K|=Q$, so $|k_L|=Q^n$. Reduction of the <field norm> on units is the finite-field norm
$$
k_L^\times\longrightarrow k_K^\times,\qquad a\longmapsto a^{1+Q+\cdots+Q^{n-1}}.
$$
The <multiplicative group of a finite field is cyclic>. The displayed power map on a group of order $Q^n-1$ has image of order $Q-1$, and is therefore onto $k_K^\times$. Given a unit $u\in\mathcal O_K^\times$, first choose $b_1\in\mathcal O_L^\times$ with $N(b_1)\equiv u\pmod\pi$.
We can correct the norm at each subsequent power of $\pi$. For $r\geq1$ and $a\in\mathcal O_L$, expanding the product over the cyclic <Galois group> gives
$$
N(1+\pi^r a)\equiv1+\pi^r\operatorname{Tr}_{k_L/k_K}(\bar a)\pmod{\pi^{r+1}}.
$$
All products involving at least two correction terms are divisible by $\pi^{2r}$, hence by $\pi^{r+1}$. The residue <field trace> is onto even if the residue characteristic divides $n$. Indeed its <polynomial> $T(X)=X+X^Q+\cdots+X^{Q^{n-1}}$ is nonzero of degree $Q^{n-1}<Q^n$; it cannot vanish at every element of $k_L$. Its values lie in $k_K$, since raising $T(a)$ to $Q$ cyclically permutes the summands. Thus it is a nonzero $k_K$-linear map to the one-dimensional space $k_K$, and is surjective.
Suppose $N(b_r)\equiv u\pmod{\pi^r}$. Choose $\bar a_r$ whose trace is the coefficient of $\pi^r$ in $u/N(b_r)-1$, and put $b_{r+1}=b_r(1+\pi^r a_r)$. Then $N(b_{r+1})\equiv u\pmod{\pi^{r+1}}$, and $b_{r+1}\equiv b_r\pmod{\pi^r}$. By <completeness> the units $b_r$ converge to a unit $b$. The <field norm> is <continuous>, being a polynomial in coordinates of multiplication, so $N(b)=u$. Multiplying such a unit by a power of $\pi$ proves
$$
\boxed{N_{L/K}(L^\times)=\{x\in K^\times:v_K(x)\equiv0\pmod n\}.}
$$
The proof uses no local class field theory, and works in either characteristic.
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