Let $L/ \Bbb{Q}_p$ be finite extension. Let $o$ be ring of integers of $L$.Let $π$ be a uniformizer, and $q$ be order of residue field.
I want to determine the order of $(o/\pi^no)^\times $. I know the answer, which is $q^n-q^{n-1}$.
I guess this is just $\phi(q^n)$($Φ$ is Euler function), but I don't know the reason why.
How can I prove $＃(o/\pi^no)^\times＝q^n-q^{n-1}$?
Let $q$ be the cardinality of the residue field of $L$. Then, for example by Proposition 5.4.5 (vii) in Gouvea's excellent book, any element in $O/\pi^n O$ can be written as a sum $$ a_0+a_1\pi+a_2\pi^2+\cdots+a_{n-1}\pi^{n-1}, $$ where $a_i\in O/\pi O$. Thus the order of $O/\pi^n O$ is $q^n$. Now suppose that a coset in $O/\pi^n O$ is a unit in $O/\pi^n O$. It is easy to see that any representative of this coset must have $a_0\neq 0$. Hence the non-units in $O/\pi^n O$ are precisely of the form $$ a_1\pi+a_2\pi^2+\cdots+a_{n-1}\pi^{n-1}. $$ In particular, there are $q^{n-1}$ of them.