Field norm restricted to groups of units in local fields 
Let $K$ be a local field (i.e. a field on which there exists a discrete additive valuation $v$). Let $O_K$ be the elements $x$ in $K$ with $v(x)\ge 0$ and let $M_K$ denote the set of elements in $K$ with $v(x)>0$.
  Let $U_K^{(s)}=1+M_K^s$ be the group of $s$-units in $O_K$, where $s\in \mathbb N$. Now
  Let $K_n$ be an unramified extension of $K$ of degree $n$, which is Galois and cyclic, ie $Gal(K_n/K)\cong \mathbb Z/n \mathbb Z$ and define the corresponding $O_{K_n}, M_{K_n}$ and $U^{(s)}_{K_n}$ in the same way (it is well known that $v$ extends uniquely to $K_n$). Show that the norm of an element in $U^{(s)}_{K_n}$ lies in $U^{(s)}_{K}$, namely $\text{Nm}_{K_n/K}(U^{(s)}_{K_n})\subset U^{(s)}_{K}$.

My only idea was to pick uniformizers $\pi_K$ and $\pi_{K_n}$ in $K$ and $K_n$ respectively. Let $x=1+\mu \pi_{K_n}^s$ from $U^{(s)}_{K_n}$, for $\mu \in O_{K_n}$ and let $\sigma$ be a generator of the Galois group.
Then $\text{Nm}(x)=(1+\mu \pi_{K_n}^s)(1+\sigma(\mu) \sigma(\pi_{K_n})^s)...(1+\sigma^{n-1}(\mu) \sigma^{n-1}(\pi_{K_n})^s)$.
Then I would use the fact that the extension is unramified, but as $s$ gets larger, the computations get messier and messier, which makes me think I am on a wrong track.
I was also curious whether this property holds even if the extension was not unramified, but say, still cyclic.
(I am using terminology of Neukirch's book from chapter 2)
 A: For a Galois extension $L/K$ of local fields, the effect of the norm map $N=N_{L/K}$ on the filtration $U_L^n$ is completely known. The best reference , I think, is Serre's book "Local Fields", chap. 5 and 14.
Chap. 5 considers an extension $L/K$ of local fields with group $G$, such that the residual field extension is separable. The unramified case is easy: if $x=1+y$, with $y \in M_L^n$, one has $\sigma (x)=1+\sigma (y)$ for all $\sigma \in G$, and $\sigma (y)\in M_L^n$, so that the product of all the $1+\sigma (y)$ is congruent to $1+$ the sum of all the $ \sigma (y)$ modulo $M_L^{2n}$ ; if $L/K$ is unramified, $M_L^n \cap K = M_K^n$, and we are done.The general case is a combination of the non ramified and totally ramified cases. It remains to solve the latter case. For this, one introduces the so-called Hasse-Herbrand function $\psi (n)$ attached to the totally ramified extension $L/K$ in order to study the effect of the map $N_n : U_L^{\psi(n)}/U_L^{\psi(n)+1} \to U_L^n/U_L^{n+1}$ induced by the norm, and one shows that $ker N_n\cong G_ \psi (n)/G_ {\psi (n)+1}$, where $G_r$ denotes the $r$-th ramification subgroup in the lower numbering. If the residual field is perfect and $N_n$ is injective, it is also surjective. If the residual field is finite, one can derive an isomorphism $\delta_n: U_K^n/U_K^{n+1}N(U_L^{\psi (n)}) \cong G_ \psi (n)/G_ {\psi(n)+1}$ .
In chap. 15 , to go further, one supposes that the residual field of $K$ is finite and $G$ is abelian. The reciprocity isomorphism of local CFT induces an isomorphism $\omega_n : U_K^n/U_K^{n+1}N(U_L^{\psi (n)}) \cong G^n/G^{n+1}$ in the upper numbering. The Hasse-Arf theorem states that $G_ \psi (n)=G^n$, and one can show the following comparison result : $\omega_n (x) = \delta_n (x^{-1})$ for all $x \in U_K^n/U_K^{n+1}N(U_L^{\psi (n)})$ ./.
