NB: Please restrict answers to hints and not solutions.
Problem: Use the theory of the Herbrand quotient $q(A)=H^{0}(A)/H^{1}(A)$ to show that, if $K$ is a finite extension of $\mathbb{Q}_p$, and $U_n$ is the set of nth roots of unity in $K$, then $|K^*/(K^*)^n|=n|U_n|/|n|_K$.
Here is my working so far. Let $G$ be a cyclic group of order $n$ generated by $\sigma$ and define automorphisms $\Delta=1-\sigma$ and $N=1+\sigma+\sigma^2+...+\sigma^{n-1}$.
Let $G$ act trivially on $K^*$. Then Hilbert's Theorem 90 says that $H^1(K^*)$ is trivial, so $q(K^*)=H^0(K^*)=|Ker(\Delta)| /|Im(N)| =K^*/(K^*)^n$. So it's enough to show that $q(K^*)=n|U_n|/|n|_K$.
Next we notice that there is a short exact sequence $0 \rightarrow U_n \rightarrow K^* \rightarrow (K^*)^n \rightarrow 0$, where the third arrow is 'take the nth power'. Here, however, I get confused. Doesn't the existence of this sequence imply (via the group isomorphism theorem) that $|K^*/(K^*)^n|=|U_n|$ without the extra factor of $n/|n|_K$ ?
I repeat that I would prefer answers to sort out this confusion with this last point, rather than giving away the answer to the problem.