Vector Calculus Identity Help Needed Ive been given a question:
Prove that if $f$ is a (smooth) scalar field and $\overrightarrow {G}$ is an irrotational vector field, then $$(∇f × \overrightarrow {G} )f$$ is solenoidal
Ive got the identities in front of me but i dont know how to apply them to this question.
I know I have to use the identity $$∇·(fG)= (∇f)·G+ f∇·G$$
Any help will be appreciated.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\nabla\cdot\bracks{\pars{\nabla f \times \vec{G}}f}
\\[5mm] = &
\overbrace{\nabla f\cdot \pars{\nabla f \times \vec{G}}}^{\ds{=\ 0}}\ +\
f\,\nabla\cdot\pars{\nabla f \times \vec{G}} =
f\bracks{\vec{G}\cdot\underbrace{\pars{\nabla\times\nabla f}}_{\ds{=\ \vec{0}}} - \nabla f \cdot \underbrace{\pars{\nabla \times G}}_{\ds{=\ \vec{0}}}} =
\color{#f00}{0}
\end{align}
