Maximal ideal $\mathfrak m$ of a non-discrete valuation ring of rank one satisfies $\mathfrak{m}^{2}=\mathfrak{m}$ I have a problem when reading Gabber's Almost Ring Theory.

Let $V$ be a non-discrete valuation ring of rank one. Does its maximal ideal $\mathfrak{m}$ satisfy $\mathfrak{m}^{2}=\mathfrak{m}$?

Here is my attempt. Take $\sum x_{i}y_{j}\in\mathfrak{m}^{2}$, $x_{i},y_{j}\in\mathfrak{m}$. Then $v(\sum x_{i}y_{j})\leq\max\{v(x_{i})v(y_{j}),...\}$, so that $v(\sum x_{i}y_{j})\leq 1$ and $\sum x_{i}y_{j}\in\mathfrak{m}$. However, this can not prove that $\mathfrak{m}\subset \mathfrak{m}^{2}$.
 A: Let $\Gamma$ be the value group of $(V,v)$. Denote the valuation field by $K$.
Suppose that there is an element $x\in V$ admitting the minimal nonzero valuation $v_{0}=v(x)$. Since the rank of $V$ is $1$, we can identify $\Gamma\otimes_{\mathbb{Z}}\mathbb{Q}$ as $\mathbb{Q}$.
For any element $y\in V$, if we put $v(y)$ in the additive group $\Gamma\otimes_{\mathbb{Z}}\mathbb{Q}$,
then we can write it as $v(y)=mv_{0}+r,$ where $m\in\mathbb{Z}, 0\leq r<v_{0}$. Consider the element $y\cdot x^{-m}$, it has valuation strictly smaller than $v_{0}$, thus it must be an unit (Notice that, by our notation, an unit has valuation $1$, but under the isomorphism $\Gamma\otimes_{\mathbb{Z}}\mathbb{Q}\simeq \mathbb{Q}$, it corresponds to $0\in\mathbb{Q}$).
This shows that $\Gamma$ is isomorphic to $\mathbb{Z}$, which is a contradiction to the fact that $V$ is non-discrete!
For each element $x\in\mathfrak{m}$, its valuation $v(x)<1$ is not minimal in $\Gamma$. So we can take $y\in \mathfrak{m}$ such that $v(y)<v(x)$, then $xy^{-1}$ must also be an element in $\mathfrak{m}$, which proves that $\mathfrak{m}=\mathfrak{m}^{2}$.
