Let $K \supseteq \mathbb{Q}_p$ be a $p$-adic field. Let $\mathcal{O}$ be the ring of integer of $K$ and $\mathfrak{m}=(\pi)$ be the maximal ideal of $\mathcal{O}$, where $\pi$ is the prime element or uniformizer of the discrete valuation ring $\mathcal{O}$. Let us consider the $p$-adic valuation (additive valuation) $v$ such that $v(\pi)=1$.
Also let $\mathcal{O}^{*}$ be the set of units of $\mathcal{O}$ defined by $$\mathcal{O}^*=\{x \in \mathcal{O} \ | \ v(x)=0 \}.$$
Question:
Suppose we consider the $3$-adic field $\mathbb{Q}_3$. Then $p=3$ is the prime element and the maximal ideal is $ \mathfrak{m}=(3)$.
Here $v(3)=v_3(3)=1$ but $v_3(3+1)=v_3(4)=0$, i.e., $p=3$ is uniformizer while $p+1$ is a unit.
Now if we se the general situation, $v(\pi)=1$ but $v(\pi+1) \neq 0$ i.e, $\pi$ is uniformizer while $\pi+1$ not unit.
Why is so ? What is the crucial part I am missing here?
Am I forgeting the extension field ?
Though in unramified extension the uniformizer doesn't change.
When $\pi+1$ will be an unit provided $\pi$ is uniformizer ?
Kindly help