If I know that $\vec \nabla \cdot \vec v=0$, can I say that:
$$( \vec v \cdot \vec \nabla )\vec v=\underbrace{(\vec \nabla \cdot \vec v)}_{=0} \vec v=0 $$ ?
Note: this is a question I asked in Physics StackExchange but as it is mainly mathematical I thought it could be relevant to post it here too.
No. Try $v := (x,-y)$. Then $\nabla \cdot v = 0$ but $(v\cdot \nabla) v = (x,y)$. Problem is $(v \cdot \nabla) v \neq (\nabla \cdot v) v$ in general.
Edit: There seems to be some confusion as to what $v \cdot \nabla$ means, especially from the comments of the OP who seem to suggest that he/she treats $v$ and $\nabla$ as vectors which could be freely multiplied together in any order.
In the above example, $v\cdot\nabla$ is the operator $x \frac{\partial}{\partial x} - y \frac{\partial}{\partial y}$ which, being an operator, does not make sense unless you apply it to something. On the other hand, $\nabla \cdot v$ means apply the operator "$\nabla \cdot$" to $v$, resulting in the scalar function $\frac{\partial x}{\partial x} + \frac{\partial (-y)}{\partial y} = 0$.