Assuming the relevant partial derivatives exist and are continuous, $\text{div}(\text{curl}(\textbf{F}))=0$, where $\textbf{F}:\mathbb{R}^3 \to \mathbb{R}^3$ is a vector field. This is easily proved by writing out the components and showing that the mixed partials all cancel (they are equal beause they are continuous).
However, one can also write $\text{div}(\text{curl}(\textbf{F}))=\nabla\cdot(\nabla \times \textbf{F})$, where $\nabla$ is the operator $\left \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle$. There is a temptation to say that $\nabla \times \textbf{F}$ is a "vector" perpendicular to $\nabla$, so $\nabla \cdot (\nabla \times \textbf{F})=0$. This is not correct since $\nabla$ is not really a vector. Is there some sense in which this is a correct justification, though? (Treating things purely algebraically?) I don't think it is rigorous enough because it relies on commutativity, i.e. being able to take partial derivatives in any order and get the same answer.