Size of units of a quotient of a discrete valuation ring

Question

Let $R$ be a discrete valuation ring and let $\mathfrak m$ be its maximal ideal. Suppose that $R/\mathfrak m$ is a finite field of characteristic $p$. Let $\# R/\mathfrak m:=p^f$ for some $f \in \mathbb N_{\geq 1}$. If we take some $r \in \mathbb N_{\geq 1}$, how do we calculate $\#(R/\mathfrak m^r)^{\times}$?

Answer 1

$\mathfrak{m}= (\pi)$ is principal.

Take $S$ a set of representatives of $R/(\pi)$ in $R$.

We have the disjoint union $$R/(\pi^r)=\bigcup_{s\in S} s+(\pi)/(\pi^r)=\bigcup_{s\in S} s+\pi \ (R/(\pi^{r-1}))$$

Whence $$|R/(\pi^r)| = |S| \ |R/(\pi^{r-1})|=|R/(\pi)|^r$$

The units are simply

$$|R/(\pi^r)^\times| = |R/(\pi^r)\ -\ (\pi)/(\pi^r)| = |R/(\pi)|^r-|R/(\pi)|^{r-1}$$