Surjectivity of norm map in totally ramified extension

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Let K be a p-adic field, L a totally ramified finite extension of K. Prove that $N_{L/K}(U^1_L)=U^1_K$ iff L/K is tamely ramified, where $U_K^1=1+\pi_K\mathcal{O}_K,U_L^1=1+\pi_L\mathcal{O}_L$ are the 1st higher unit groups.

I believe this problem has something to do with class field theory, but I had little grasp on that subject... Any help would be appreciated.

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Since it seems that you may like it, I'll expand on reuns's comment.

We need only show the equality stated in the comment. Take $\lambda \in O_{K}$, we want to find $x\in O_{K}$ such that $$(1+\pi_{K}x)^{n}=1+\pi_{K}\lambda$$ is satisfied.

By looking at the development of the left hand side, this is equivalent to find an $x\in O_{K}$ such that the equation $$\lambda=x(n+x\pi_{K}\binom{n}{2}+...+x^{n-1}\pi^{n-1})$$ is satisfied.

But by going to the residue field this becomes: $[\lambda]=[n]x$, which has always a solution in $O_{K}$ thanks to Hensel's lemma which we can apply because $p$ does not divide $n$.