Proof That a Cross Product of Gradients is Solenoidal

Theorem
If  
\[f, \;\]
  are differentiable functions then  
\[(\mathbf{\nabla}f) \times (\mathbf{\nabla}g)\]
  is solenoidal, so that  
\[\mathbf{\nabla} \cdot(\mathbf{\nabla}f) \times (\mathbf{\nabla}f)=0\]
  (1). Proof
Use the identity  
\[\mathbf{\nabla} \cdot (\mathbf{a} \times \mathbf{b})=\mathbf{b} \cdot (\mathbf{\nabla} \times \mathbf{a})- \mathbf{a} \cdot (\mathbf{\nabla} \times \mathbf{b}) \]
  so that (1) becomes
\[\mathbf{\nabla} \cdot ((\mathbf{\nabla} f \times \mathbf{\nabla} g) =\mathbf{\nabla} g \cdot (\mathbf{\nabla} \times \mathbf{\nabla} f) - \mathbf{\nabla} f \cdot (\mathbf{\nabla} \times \mathbf{\nabla} g)\]

But  
\[\mathbf{\nabla} \times \mathbf{\nabla} f = \mathbf{\nabla} \times \mathbf{\nabla} g =0\]

Hence  
\[\mathbf{\nabla} \cdot(\mathbf{\nabla}f) \times (\mathbf{\nabla}f)=0\]