I have Maxwell's equations as follows:
$$\nabla \times \mathbf{E} = - \dfrac{\partial}{\partial{t}}\mathbf{B} \ \ \ \ \text{Faraday's law}$$ $$\nabla \times \mathbf{H} = \mathbf{J} + \dfrac{\partial{\mathbf{D}}}{\partial{t}} \ \ \ \ \text{Ampére's law}$$ $$\nabla \cdot \mathbf{D} = \rho \ \ \ \ \text{Gauss's law}$$ $$\nabla \cdot \mathbf{B} = 0 \ \ \ \ \text{Gauss's law}$$ where $\mathbf{E}$ is the electric field (V/m), $\mathbf{H}$ is the magnetic field (A/m), $\mathbf{D}$ is the electric displacement flux density (C/m$^2$), and $\mathbf{B}$ is the magnetic flux density (Vs/m$^2$ or Webers/m$^2$).
I am told that Gauss's law $\nabla \cdot \mathbf{B} = 0$ is derivable from Faraday's law $\nabla \times \mathbf{E} = - \dfrac{\partial}{\partial{t}}\mathbf{B}$ by taking the divergence and noting that $\nabla \cdot ( \nabla \times \mathbf{E}) = 0$ for any vector $\mathbf{E}$. Why/how is $\nabla \cdot ( \nabla \times \mathbf{E}) = 0$, and how do we then get $\nabla \cdot \mathbf{B} = 0$ from $\nabla \times \mathbf{E} = - \dfrac{\partial}{\partial{t}}\mathbf{B}$?