= Solution
We give the full local-field norm-index argument. Let $G=\langle\sigma\rangle$ be cyclic of order $n$. On an additive $G$-<module> set $D=\sigma-1$ and $N=1+\sigma+\cdots+\sigma^{n-1}$. The two cyclic <Tate cohomology> <groups> are
$$
\widehat H^0(G,M)=M^G/NM,\qquad\widehat H^{-1}(G,M)=\ker N/DM.
$$
When these are finite, the <Herbrand quotient> is $h_G(M)=|\widehat H^0(G,M)|/|\widehat H^{-1}(G,M)|$. The periodic complex alternating $D$ and $N$ yields a six-term periodic exact sequence from a short exact sequence of <modules>. Taking alternating products of orders in that exact sequence proves
$$
h_G(M)=h_G(M')h_G(M'')\qquad(0\to M'\to M\to M''\to0).
$$
For a finite <module>, $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$ show that the two <Tate cohomology> <groups> have equal order. Thus the <Herbrand quotient of a finite module> is one. For the trivial integral <module> $\mathbb Z$, the two groups are $\mathbb Z/n\mathbb Z$ and zero, giving $h_G(\mathbb Z)=n$.
We also need <cyclic cohomology of a regular lattice>. In $R[G]$, fixed coefficient vectors are constant, and every such vector is a norm of a basis vector times its coefficient. The <kernel> of $N$ consists of vectors whose coefficient sum is zero. For such a vector $(a_i)$, solve $b_{i-1}-b_i=a_i$ successively around the cycle; the final equation holds precisely because $\sum_i a_i=0$. Thus $\ker N=DR[G]$. Both <Tate cohomology> <groups> vanish, so $h_G(R[G])=1$, for $R=\mathbb Z$ or $\mathbb Z_p$, and likewise for finite direct sums.
Now take a cyclic extension $L/K$ of <p-adic fields> of degree $n$. Write $U_L^r=1+\mathfrak p_L^r$. For $r>v_L(p)/(p-1)$, the convergent <p-adic logarithm> and <p-adic exponential> give mutually inverse $G$-equivariant <group isomorphisms>
$$
U_L^r\cong\mathfrak p_L^r,
$$
where the right side is additive. The convergence assertion follows from the estimates $v_L(x^j/j)\ge jv_L(x)-v_L(p)v_p(j)$ and $v_L(x^j/j!)\ge jv_L(x)-v_L(p)j/(p-1)$; formal inverse identities therefore hold on this ball.
By the <normal basis theorem>, choose $\theta$ so that $\{\sigma^i\theta\}$ is a $K$-<basis> of $L$. The lattice $A=\sum_i\mathcal O_K\sigma^i\theta$ is a regular $\mathcal O_K[G]$-<module>. It is a finite direct sum of regular $\mathbb Z_p[G]$-<modules>, hence its <Tate cohomology> vanishes. The lattices $A$ and $\mathfrak p_L^r$ are commensurable: their intersection has finite index in both. The exact sequences with these finite quotients establish finiteness of the <Tate cohomology> <groups> and show $h_G(\mathfrak p_L^r)=1$. Thus $h_G(U_L^r)=1$. Since $\mathcal O_L^\times/U_L^r$ is finite, $h_G(\mathcal O_L^\times)=1$ as well.
The $G$-equivariant <valuation> exact sequence is
$$
1\longrightarrow\mathcal O_L^\times\longrightarrow L^\times\xrightarrow{v_L}\mathbb Z\longrightarrow0.
$$
Its last term has trivial $G$-action. The multiplication rule gives the <Herbrand quotient of the local multiplicative group>:
$$
h_G(L^\times)=n.
$$
All the <Tate cohomology> <groups> here are finite by the lattice argument and the exact sequence; finiteness of the norm index has not been assumed.
For completeness we prove the needed <Hilbert theorem 90>. If $a\in L^\times$ has <field norm> one, put $A_0=1$ and $A_i=\prod_{j=0}^{i-1}\sigma^j(a)$. The linear map $c\mapsto b=\sum_{i=0}^{n-1}A_i\sigma^i(c)$ is not zero by <linear independence of distinct field embeddings>. Choose $c$ with $b\ne0$. The relation $A_n=1$ gives $a\sigma(b)=b$, so $a=b/\sigma(b)$. This proves $\widehat H^{-1}(G,L^\times)=0$. The zeroth <Tate cohomology> <group> is $K^\times/N_{L/K}L^\times$. Consequently the <cyclic local norm index> is
$$
\boxed{[K^\times:N_{L/K}L^\times]=[L:K]=n.}
$$
Writing $n=ef$, the <valuation> of a <field norm> is $v_K(Nx)=f v_L(x)$. The valuation quotient has order $f$, and a norm with <valuation> zero comes from a unit. Therefore
$$
\boxed{[\mathcal O_K^\times:N_{L/K}\mathcal O_L^\times]=e.}
$$
In particular every unit is a norm in an <unramified extension>. These results give the norm assertions used in the <quadratic Hilbert symbol> computation.
We finish with the <Hasse norm theorem>: \b[for a cyclic extension of number fields, an element is a global norm if and only if it is a norm at every place]. More precisely the local norm is the norm from the product $L\otimes_K K_v$; in a <Galois extension> its factors have the same norm subgroup. The forward implication is immediate. For the converse associate to $a\in K^\times$ the <cyclic algebra> $(L/K,\sigma,a)$. The <splitting criterion for a cyclic algebra>, proved below, identifies its splitting with $a\in N_{L/K}L^\times$. At a place where the tensor product of fields has several factors, this <algebra> is a matrix algebra over the <cyclic algebra> of one decomposition-field factor, with the generator restricted to its cyclic decomposition subgroup. Thus the local norm assumption makes its <Brauer class> zero at every place. Injectivity in the <Albert-Brauer-Hasse-Noether theorem> makes the global <Brauer class> zero, giving the desired global norm. This is the required proof outline; cyclicity is essential to this reduction.
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