$\nabla \cdot (\nabla \times \vec A) = 0 $ Proof Show that $\nabla \cdot (\nabla \times \vec A) = 0  $ for an arbitrary differentiable vector field $\vec A$ using Stokes' Theorem for an arbitrary closed surface S followed by Gauss' Theorem.
An attempt at the solution:

Using Stokes' Theorem:
$$\oint_{C=\partial S} \vec A \cdot d\vec l = \int_S (\nabla \times \vec A) \cdot d\vec S$$
Letting $S=\partial V$ and using Gauss' Theorem:
$$\oint_{S=\partial V} (\nabla \times \vec A) \cdot d\vec S = \int_V \nabla \cdot (\nabla \times \vec A) \:dV  $$
Hence,
$$\oint_{C=\partial S} \vec A \cdot d\vec l =\int_V \nabla \cdot (\nabla \times \vec A) \:dV $$
It looks like I should be able to arrive at the solution from some logic being applied to this equality, but I am unsure what that would be.
 A: Okay. To start, we take a closed curve $C$ on the boundary of V ($S = \partial V)$, so that $C\subset \partial V$. Then the surface $\partial V$ is broken into two parts, suppose they are $S_1, S_2$. ($S_1\cup S_2=S$)
Then, as your attempted solution; using Stoke's Theorem,
$$\oint_C \vec{A} \cdot d\vec{l}=\int_{S_1}(\nabla \times \vec{A})\cdot d \vec{S}$$
$$-\oint_C \vec{A} \cdot d\vec{l}=\int_{S_2}(\nabla \times \vec{A})\cdot d \vec{S}$$
I don't know how I should formally state this, but since the orientation of the curve $C$ in the LHS of the first and second integral are different, they have the same absolute value and different signs.
So add these, so that
$$0=\int_{S_1}(\nabla \times \vec{A})\cdot d \vec{S}+\int_{S_2}(\nabla \times \vec{A})\cdot d \vec{S}=\oint_{S}(\nabla \times \vec{A})\cdot d \vec{S}$$
then now you can use Gauss' Theorem, as you did in your solution.
Then, we get
$$\oint_{S}(\nabla \times \vec{A})\cdot d \vec{S}=\int_V \nabla \cdot (\nabla \times\vec{A})dV=0$$
Since this must hold for arbitrary $\vec{A}$,
$$\nabla \cdot (\nabla \times\vec{A})=0$$
I hope this helped.
A: \begin{align}
  & \operatorname{curl}\,A=\nabla \times F=\left| \begin{matrix}
   i & j & k  \\
   \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}  \\
   P & Q & R  \\
\end{matrix} \right| \\ 
 & \operatorname{curl}\,A=\left( \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z} \right)i+\left( \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x} \right)j+\left( \frac{\partial Q}{\partial X}-\frac{\partial P}{\partial y} \right)k \\ 
 & \nabla .(\operatorname{curl}\,A)=\operatorname{div}(\operatorname{curl}\,A)=\frac{{{\partial }^{2}}R}{\partial x\partial y}-\frac{{{\partial }^{2}}Q}{\partial x\partial z}+\frac{{{\partial }^{2}}P}{\partial y\partial z}-\frac{{{\partial }^{2}}R}{\partial y\partial x}+\frac{{{\partial }^{2}}Q}{\partial z\partial x}-\frac{{{\partial }^{2}}P}{\partial z\partial y}=0 \\ 
\end{align}
we suppose $P$ , $Q$ and $R$ are continues
