Identity for Product of a Function With a Vector Field Satisfying Poisson's Equations

Theorem
If  
\[f\]
  is a function and  
\[\mathbf{v}\]
  is a vector field satisfying  
\[\nabla^2 (f \mathbf{v}) =0\]
  and  
\[\mathbf{\nabla} \times (\mathbf{\nabla} \times (f \mathbf{v}))=0\]
  then  
\[\mathbf{\nabla} ((\mathbf{\nabla} f) \cdot \mathbf{v} + f \mathbf{\nabla} \cdot (\mathbf{v})) =0\]

Proof
\[\mathbf{\nabla} \times (\mathbf{\nabla} \times (f \mathbf{v}))= \mathbf{\nabla} (\mathbf{\nabla} \cdot (f \mathbf{v}))- \nabla^2 (f \mathbf{v})\]

\[\nabla^2 (f \mathbf{v}) =\mathbf{\nabla} \times (\mathbf{\nabla} \times (f \mathbf{v}))=0\]
&nbsp so  
\[\mathbf{\nabla} (\mathbf{\nabla} \cdot (f \mathbf{v} ))=0 \]

Now use the identity  
\[\mathbf{\nabla} \cdot (f \mathbf{v}) =(\mathbf{\nabla} f ) \cdot \mathbf{v} + f ( \mathbf{\nabla} \cdot \mathbf{v})\]
  to obtain  
\[\mathbf{\nabla} ((\mathbf{\nabla} f ) \cdot \mathbf{v} + f \mathbf{\nabla} \cdot \mathbf{v}) =0\]