If
\[\mathbf{v}\]
and \[\mathbf{w}\]
are vectors then \[(\mathbf{v} \times \mathbf{\nabla}) \cdot \mathbf{w} = \mathbf{v} \cdot (\mathbf{\nabla} \times \mathbf{w})\]
Proof
Let
\[\mathbf{v}=v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k} \]
and \[\mathbf{w}=w_1 \mathbf{i} + w_2 \mathbf{j} + w_3 \mathbf{k} \]
\[\begin{equation} \begin{aligned} (\mathbf{v} \times \mathbf{\nabla}) \cdot \mathbf{w} &=
((v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k}) \times (\frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k} +))\cdot (w_1 \mathbf{i} + w_2 \mathbf{j} + w_3 \mathbf{k}) \\ &=
((v_2 \frac{\partial}{\partial z} - v_3 \frac{\partial}{\partial y})\mathbf{i} + (v_3 \frac{\partial}{\partial x} - v_1 \frac{\partial}{\partial z})\mathbf{j} + (v_1 \frac{\partial}{\partial y} - v_2 \frac{\partial}{\partial x})\mathbf{k}) \cdot (w_1 \mathbf{i} + w_2 \mathbf{j} + w_3 \mathbf{k}) \\ &=
(v_2 \frac{\partial}{\partial z} - v_3 \frac{\partial}{\partial y})w_1 + (v_3 \frac{\partial}{\partial x} - v_1 \frac{\partial}{\partial z})w_2 + (v_1 \frac{\partial}{\partial y} - v_2 \frac{\partial}{\partial x})w_3 \\ &=
v_2 \frac{\partial w_1}{\partial z} - v_3 \frac{\partial w_1}{\partial y} + v_3 \frac{\partial w_2}{\partial x} - v_1 \frac{\partial w_2}{\partial z} + v_1 \frac{\partial w_3}{\partial y} - v_2 \frac{\partial w_3}{\partial x} \\ &=
v_1 (\frac{\partial w_3}{\partial y} - \frac{\partial w_2}{\partial z}) + v_2 (\frac{\partial w_1}{\partial z} - \frac{\partial w_3}{\partial x}) + v_3 (\frac{\partial w_2}{\partial x} - \frac{\partial w_1}{\partial y}) \\ &=
\mathbf{v} \cdot (\mathbf{\nabla} \times \mathbf{w} ) \end{aligned} \end{equation}\]