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Proof That The Divergence of the Curl of a Vector is Zero

The Divergence Theorem states that for any vector field  
\[\mathbf{A}\]
  with differentiable components defined on a volume  
\[V\]
  with boundary  
\[S\]

\[\int \int \int_V \mathbf{\nabla} \cdot \mathbf{A} dV = \int \int_S \mathbf{A} \cdot \mathbf{n} dS \]

Let  
\[\mathbf{A} = \mathbf{\nabla} \times \mathbf{F}\]
  for some vector field  
\[\mathbf{F}\]

Then  
\[\int \int \int_V \mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{F}) dV = \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS \]

Then the right hand side is zero. Hence  
\[\int \int \int_V \mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{F}) dV =0 \]

The surface is arbitrary and so is the volume, hence  
\[\mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{F}) =0\]