$K$ is a field
I have to show the existence of a unit $u$, i.e. an element where the constant part of the formal power series is not $0$ such that $f=ux^n$, i.e $x^n=fu$
From the excercise sheet I cannot tell if this should hold for every $n\in\mathbb{N}$ or just one $n\in\mathbb{N}$. Is it possible to sho it for every $n\in\mathbb{N}$ ?
How can I start?
Assume it would hold for different $n$. So there exist $u,v$ units in $K[[x]]$ such that $$ u x^n = f = v x^m $$ with $m,n\in \mathbb{N}$ and $n\neq m$. Wlog assume $n>m$, then we have $$ 0 = vx^m - u x^n = vx^m(1- uv^{-1} x^{n-m)}. $$ As $K[[x]]$ is an integral domain and $ux^n \neq 0$ we get $1-uv^{-1} x^{n-m}=0$, thus $x^{n-m}$ is a unit, which is a contradiction. So, indeed, for every $f$ there is a unique $n$ with the desired property.
To show the general statement it might be helpful to first trying to solve the case $$ f= \sum_{n\geq 1} x^n. $$ From here you should get the intuition how to proceed.