I know that the $v: f \mapsto -\deg(f)$ is a discrete valuation on the field of complex rational functions $\mathbb{C}(X)$ (the quotient field of $\mathbb{C}[X]$). The valuation ring $\mathcal{O}_v$ is a local PID with a single maximal ideal $m_v$. Furthermore, $m_v$ is generated by any element $f \in \mathbb{C}(X)$ with valuation $v(f) = 1$.
For this concrete valuation, we can see that $m_v$ contains (besides $0$) only elements $\frac{f}{g}$ with (relatively prime) $f, g \in \mathbb{C}[X]$, $\deg(f) < \deg(g)$ and $g \neq 0$. $\mathcal{O}_v$ consists of $m_v$ along with the units, which are exactly the elements $e \in \mathbb{C}(X)$ with valuation $0 = v(e) = -\deg(e)$, i.e. $e \in \mathbb{C} \setminus \{ 0 \}$.
Therefore, we can take e.g. $f = \frac{1}{X}$ and conclude $m_v = (f)$. But now I don't see how it is possible to write $\frac{1}{X-1} \in m_v$ in the form $e f^n$ where $e \in \mathcal{O}_v^\times = \mathbb{C} \setminus \{ 0 \}$ and $n \in \mathbb{N}_0$.
What is wrong with my reasoning? How can $m_v$ be generated by $\frac{1}{X}$?
The units in $\mathcal{O}_v$ are not only the units in $\mathbb{C}$, but all degree $0$ elements of $\mathbb{C}(X)$.
In particular, $\frac{X}{X-1} \in \mathcal{O}_v^{\times}$.
Note: for a rational function $f/g$ with $f,g$ polynomials you define $\deg(f/g) = \deg(f) - \deg(g)$.