Let $F$ be a non-archimedian local field with valuation $\nu$. Then $\mathcal{O}=\{x\in F: \nu(x)\geq 0\}$ is the ring of integers of $F$. $\mathfrak{m}=\{x\in F: \nu(x)> 0\}$ is the maximal ideal of $\mathcal{O}$. Let $\pi \in \mathcal{O}$ such that $\nu(\pi)=1$. It is said that every element $x$ in $F$ is of the form $\pi^n u$, where $n = \nu(x)$, $u$ is invertible in $F$. Why elements in $F$ are of the form $\pi^n u$?
In http://en.wikipedia.org/wiki/Local_field, the ring $\mathcal{O}$ is given by $\{a \in F: \nu(a) \leq 1\}$. It is different from the definition in the first paragraph which is from the book Local filed by Serre. I am confused.
Thank you very much for your help.
Regarding the confusion: The $p$-adic value $|x|_p$ is obtained from the $p$-adic valuation $\nu_p(x)$ per $|x|_p = p^{-\nu_p(x)}$ (and $|0|_p=0$ to avoid the infinity occuring in $\nu_p(0)=\infty$). So we have $\nu_p(x)\ge 0\iff |x|_p\le 1$.
Since $\nu(x)\in\mathbb N_0$ for all $x\in \mathcal O$, the number $u:=\frac{x}{\pi^{\nu(x)}}\in F$ has the property $\nu(u)=0$. Since $\mathcal O\setminus\mathfrak m=\mathcal O^\times$ this precisely says that $u$ is a unit.