= Solution
We give the detailed norm-index argument for a cyclic extension $L/K$ of <p-adic fields>; thus $K$ is a finite extension of $\mathbb Q_p$. Write $G=\langle\sigma\rangle$, $|G|=n$. For an additive $G$-<module> $M$, put $D=\sigma-1$ and $N=1+\sigma+\cdots+\sigma^{n-1}$. The two periodic <Tate cohomology of a cyclic group> <groups> and the <Herbrand quotient> are
$$
\widehat H^0(G,M)=\ker D/NM,\qquad
\widehat H^{-1}(G,M)=\ker N/DM,\qquad
h_G(M)=\frac{|\widehat H^0(G,M)|}{|\widehat H^{-1}(G,M)|},
$$
when these <groups> are finite. For multiplicative <modules>, $N$ is the product of conjugates and $D(z)=\sigma(z)/z$.
The periodic complex alternating $D$ and $N$ gives a six-term <exact sequence> for every <short exact sequence> of <modules>. Taking the alternating product of the <finite group> orders in that <exact sequence> proves
$$
h_G(M)=h_G(M')h_G(M'')\quad\text{if }0\to M'\to M\to M''\to0.
$$
For a finite <module>, the counting fibers gives $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$, so the two Tate <groups> have equal order. This proves the <Herbrand quotient of a finite module> is one. It follows that replacing a <module> by a finite-index submodule does not change its quotient, whenever the <cohomology groups> are finite.
Two basic calculations drive the norm computation. For the trivial <module> $\mathbb Z$, $D=0$ and $N$ is multiplication by $n$, so $h_G(\mathbb Z)=n$. For a regular lattice $R[G]$, with $R=\mathbb Z_p$, invariants are constant coefficient vectors and every such vector is a norm: put its constant coefficient in just one coordinate before applying $N$. The <group kernel> of $N$ consists of the coefficient vectors of sum zero. Successively taking <partial sums> solves the cyclic difference equation $Dx=y$ for every such vector $y$. Thus both Tate <groups> vanish and $h_G(R[G])=1$. This is <cyclic cohomology of a regular lattice>.
Let $U_L=\mathcal O_L^\times$ and $U_L^r=1+\mathfrak P_L^r$. For sufficiently large $r$, the <p-adic logarithm> and <p-adic exponential> are inverse $G$-equivariant <isomorphisms>
$$
\log:U_L^r\xrightarrow{\sim}\mathfrak P_L^r.
$$
To make the range explicit, $r>v_L(p)/(p-1)$ suffices: the higher terms in the logarithm and exponential have greater <valuation> than their leading terms, their series converge there, and the identities $\log(\exp x)=x$ and $\exp(\log(1+x))=1+x$ remain valid. Thus a deep multiplicative <unit group> becomes an additive $\mathbb Z_p$-lattice in $L$.
By the <normal basis theorem>, $L\cong K[G]$ as a $K[G]$-module. As a $\mathbb Q_p[G]$-module it is therefore $\mathbb Q_p[G]^{[K:\mathbb Q_p]}$. Transport the regular lattice $M_0=\mathbb Z_p[G]^{[K:\mathbb Q_p]}$ into $L$. Both $M_0$ and $\mathfrak P_L^r$ are full $G$-stable $\mathbb Z_p$-lattices, so their intersection has finite index in each. The <exact sequences> with finite quotients show that the Tate <groups> of $\mathfrak P_L^r$ are finite and that
$$
h_G(U_L^r)=h_G(\mathfrak P_L^r)=h_G(M_0)=1.
$$
Since $U_L/U_L^r$ is finite, the exact-sequence formula for the <Herbrand quotient> gives $h_G(U_L)=1$. Finally, the <valuation> sequence
$$
1\longrightarrow U_L\longrightarrow L^\times\xrightarrow{v_L}\mathbb Z\longrightarrow0
$$
has trivial $G$-action on $\mathbb Z$. Hence the <Herbrand quotient of the local multiplicative group> is
$$
h_G(L^\times)=h_G(U_L)h_G(\mathbb Z)=n.
$$
For <completeness>, the denominator vanishes by <Hilbert theorem 90>. If $N_{L/K}a=1$, put $c_0=1$ and $c_j=a\sigma(a)\cdots\sigma^{j-1}(a)$. Independence of distinct <field automorphisms> lets us choose $t\in L$ with $b=\sum_{j=0}^{n-1}c_j\sigma^j(t)\ne0$. Since $c_n=1$, shifting the sum gives $\sigma(b)=b/a$. Thus $a=b/\sigma(b)$, a multiplicative coboundary. Therefore $\widehat H^{-1}(G,L^\times)=1$. The invariant <subgroup> of $L^\times$ is $K^\times$, so
$$
\boxed{[K^\times:N_{L/K}L^\times]=[L:K]=n.}
$$
This proves the <cyclic local norm index> without assuming local reciprocity. If the extension has <ramification index> $e$ and <residue degree> $f$, the <valuation> formula $v_K(Nz)=f\,v_L(z)$ shows that the <valuation> contribution to the index is $f$. A norm is a unit only when $z$ is a unit, so the remaining contribution is
$$
[U_K:N_{L/K}U_L]=e.
$$
In particular, unit norms are surjective in an unramified cyclic extension. The norm <subgroup> is open: on deep units its logarithm is $\operatorname{Tr}_{L/K}(\mathfrak P_L^r)$, a nonzero full lattice in $K$, and therefore contains a sufficiently deep principal-unit <group>. This connects the index calculation to the open finite-index norm <subgroups> classified by <local class field theory>.
To finish, the <Hasse norm theorem> for a cyclic extension of <number fields> states that a nonzero element is a global norm if and only if it is a norm at every finite and infinite completion. Here is its Brauer-theoretic proof outline. Associate to $a\in K^\times$ the <cyclic algebra> $A=(L/K,\sigma,a)$, with relations $z\ell=\sigma(\ell)z$ and $z^n=a$. The <splitting criterion for a cyclic algebra> says that it splits exactly when $a=Nc$: if $a=Nc$, multiplication by $L$ and the operator $c\sigma$ realize it as $\operatorname{End}_K(L)$; conversely, in a split <matrix representation>, the underlying $n$-dimensional space is one-dimensional over $L$, so $z$ must act as $c\sigma$ and $z^n=Nc$.
Apply this criterion at every completion, using the restricted cyclic character and any one completion of $L$ above that place. If $a$ is a local norm everywhere, every localized Brauer class is zero. <Injectivity> in the <Albert-Brauer-Hasse-Noether theorem> forces $[A]=0$ in $\operatorname{Br}(K)$; the global splitting criterion gives $a=Nc$. Conversely, a global norm is a local norm after completion, since the norm on $L\otimes_K K_v$ is the product of the component norms, whose images coincide in the Galois case. \b[The cyclic hypothesis is essential]; the stated local-global principle does not hold for arbitrary extensions.
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