Assume $F\supset k$ is a functional field. Assume $R\subset F$ is a discrete valuation ring with a quotient filed $F$, that contains the field of constants $k$. Assume $t$ is a local parameter for $R$. Then each $z\in F^*$ admits the representation $$z=\frac{r_1}{r_2}=\frac{ut^m}{vt^n}$$ for some $r_1,r_2\in R$ and $u,v\in R^*$
Define $\nu:F\to\mathbb R\cup\{\infty\}:z=\frac{ut^m}{vt^n}\to \nu(z)=m-n$. I am told that $\nu[F^*]=\mathbb Z$. I see that this is so exactly when the parameter $t$ is not nillpotent.
Would anyone shed some light?
As pointed in the comments, a discrete valuation ring is an integral domain. If $t\not =0$ was nillpotent and $n$ was the smallest natural such that $t^n=0$ then $t^{n-1}t=0$ implies one of $t^{n-1}$ and $t$ is zero, a contradiction.
Furthermore, the sequence $t, t^2,t^3,\dots$ may only stabilize if $t=1$, which is not the case. Therefore, $\nu[F^*]=\mathbb Z$ q.e.d.