Let $R$ be a $p$-adically complete local ring with valuation $v_R$ and maximal ideal $\mathfrak{m}_R = \{ x \in R \mid v_R(x) > 0\}$. Then we have its quotient ring $R/p^n$.
How can I show that $\mu_{p^\infty}(R/p^n) = (1 + \mathfrak{m}_R)/p^n$?
Here for a ring $S$, I mean $\mu_{p^\infty}(S) = \varinjlim_{k \geq 1} \mu_{p^k}(S) = \{ x \in S \mid \exists k \geq 1 \text{ such that } x^{p^k} = 1\}$.
My attempt: For '$\supset$': suppose $1 + x \in (1 + \mathfrak{m}_R)/p^n$. Then $v_R(x) > 0$ and we can define $N := \max(\log_p\left(\frac{v_R(p^n)}{v_R(x)}\right),n)$. Then for $k \in \mathbb{Z}_{\geq N}$ we have $p^k \geq p^N \geq \frac{v_R(p^n)}{v_R(x)}$ hence $v_R(x^{p^k}) \geq v_R(p^n)$ which means $x^{p^k} = a p^n$ for some $a \in R$ and $x^{p^k} = 0$ in $R/p^n$.
Also we have in $R/p^n$ that $(1+x)^{p^k} = 1 + x^{p^k}$ because $k \geq n$, so $(1+x)^{p^k} = 1 + x^{p^k} = 1$ in $R/p^n$, which means $1+x \in \mu_{p^k}(R/p^n) \subset \mu_{p^\infty}(R/p^n)$.
For '$\subset$': suppose $x \in \mu_{p^\infty}(R/p^n)$, so $x^{p^k} = 1$ for some $k \geq 1$. We may assume that $k \geq n$. Then $(x-1)^{p^k} = x^{p^k} -1 = 0$ in $R/p^n$ which means $(x-1)^{p^k} = a p^n$ for some $a \in R$. Then $v_R(x-1)= \frac{v_R( a p^n)}{p^k} \geq \frac{n}{p^k} v_R(p) > 0$ so $x-1 \in \mathfrak{m}_R$, which means $x = 1 + (x-1) \in 1 + \mathfrak{m}_R$.
Here’s how I would do it. For $\subset$, $R/m_R$ is a field of characteristic $p$, so only $1$ is a $p^n$-th root of unity for any $n$. Thus $\mu_{p^{\infty}}(R/(p^n))$ must map to $1$ under reduction mod $m_R$, QED.
For $\supset$, normalize $v_R$ so that $v_R(p)=1$. Let $x =1+y \in 1+m_R$. Then $\Phi_{p^k}(x)=\Phi_p(x^{p^{k-1}})$ is congruent to $\Phi_p(1)=p$ mod $y$, and therefore $v_R(\Phi_{p^k}(x))\geq\min(1,v_R(y))$.
Therefore, if $k \geq 1$, $v_R(x^{p^k}-1) =v_R(y)+ \sum_{1 \leq l \leq k}{v_R(\Phi_{p^l}(x))} \geq v_R(y)+k\min(1,v_R(y))$.
So for large enough $k$, $x^{p^k}$ is congruent to $1$ mod $p^n$, and $x \in \mu_{p^{\infty}}(R/(p^n))$.