Let $m\geq 1$ and $K$ be a local field containing the $m$-roots of $1$ (I am unsure to what extent this hypothesis is relevant), and denote by $v$ a normalized valuation of $K$ given by a uniformizer $\pi$. Denote by $\mathfrak o$ its ring of integers, and $k$ the residue field. The claim I would like to show is the following
The subgroup $(K^*)^m$ of $K^*$ has index $\frac{m^2}{|m|_v}$ where $|m|_v$ is the normed absolute value of $m$ at $v$.
I have difficulties writing down a proof of this statement. Moreover, I feel like my work is contradicting the result in some case. Here is what I have done so far.
A general element $x$ of $K^*$ can be uniquely written as $x=u\pi^{v(x)}$ with $u$ a unit. Because we are interested in computing the index of $(K^*)^m$, we may assume that $v(x)=0,\ldots,m-1$. To understand the structure of the quotient, we need to study how the unit $u$ reduces inside, that is what can we say of $\mathfrak o^*/(\mathfrak o^*)^m$.
In the case where $m$ is a unit in $K$, writing down $p$ the characteristic of $k$ and $q=p^f$ its order, we have $(p,m)=(q,m)=1$. Then every non-zero element of $k$ is an $m$-th power (this can be seen using the fact that $k^*$ is cyclic). Because the polynomial $X^m-u$ is separable in $k[X]$ for any unit $u$, Hensel's lemma then implies that it has a root in $\mathfrak o$, that is in fact $\mathfrak o^* = (\mathfrak o^*)^m$. Thus, the quotient $K^*/(K^*)^m$ is generated by the image of $\pi$ and has order $m$ in this case (even though $|m|_v = 1$ since $m$ is a unit, so this seems to contradict the above statement, unless I am very confused... ?)
In the case where $m$ is not a unit, I fail to see how to proceed. Maybe we can write $m=p^am'$ with $m'$ prime to $p$ and reduce to the case where $m$ is a power of $p$.
Would somebody be able to give me a hand with this ?
The problematic step in your argument is when you conclude from $(p,m)=1$ that every element of $k$ is an $m$-th power, which is not true. As an example, consider $K={\mathbb Q}_p$, $k={\mathbb F}_p$ and $m = p-1$, in which case $(p,m)=1$ but $x^m=1$ for all $x\in k$. The point is that $k^\times$ is cyclic of order $q-1$, not $q$.
Your use of Hensel's Lemma for $(p,m)=1$ is correct, though, and shows $({\mathscr O}^{\times}:({\mathscr O}^{\times})^m)=(k^\times:(k^{\times})^m)=|\mu_m(k)|$, so $(K^\times:(K^{\times})^m)=m |\mu_m(k)|$ for $(p,m)=1$ (which is the same as $m^2 / \|m\|_K$ because you assumed $|\mu_m(k)|=m$ and because $\|m\|_K=1$).