I have been doing some revision on local field theory and have gathered up a collection of questions which I have been unable to make much progress with; there will be a few similar queries along with this problem, I hope that's ok. I wish to prove the following:
Suppose $K$ is a field that is complete with respect to a discrete valuation $v$, and $K$ has finite residue field $k$, say. Prove that when $\operatorname{char}k \nmid n$, it holds that $|K^*/(K^*)^n| = n|\mu_n(K)|$, where $\mu_n(K)$ is the group of $n$-th roots of unity in $K$.
I think this is one of the easier problems I am struggling with. I think notation and terminology is largely standard here, ask if not.
Now intuitively I can sort of see how this all ties together; working in $K^*$ modulo $(K^*)^n$ gives us a collection of cosets $a_i(K^*)^n$ and of course every one of these cosets will have order $m \leq n$ (and indeed $m \mid n$); so it's possible the right hand side of the equality is simply the product of the size of a coset (i.e. the size of $(K^*)^n$ ) and the number of such cosets.
I'm sure $\mu_n$ will come in in some obvious sense because generally if some property is true for $a$ (with regards to its $n$-th power) then it will probably also be true for $\zeta_n a$ whenever $\zeta_n \in \mu_n$. However, I'm afraid I've been unable to tie it all together. It is easy to see that the valuation of any element of $(K^*)^n$ must be divisible by $n$, but we haven't really used the residue field at all yet. If someone would be so kind as to help explain the steps I should be attempting to prove this (and ideally also explain why it makes sense to do this, if that isn't clear) then I'll be happy to give it a go myself.
You can assume I am probably familiar with most basic results on valuations, residue fields etc, I just don't know how to approach the problem. Many thanks as always.