Working in Euclidean $\RR^3$, consider a 1-form \begin{equation} F = F_x \,dx + F_y \,dy + F_z \,dz \end{equation} What is $dF$? We have \begin{align} d(F_x\,dx) &= dF_x\wedge dx \nonumber\\ &= \left( \Partial{F_x}{x}\,dx+\Partial{F_x}{y}\,dy+\Partial{F_x}{z}\,dz\right) \wedge dx \nonumber\\ &= \Partial{F_x}{z}\,dz\wedge dx - \Partial{F_x}{y}\,dx\wedge dy \end{align} Adding up three such terms, we obtain \begin{align} dF &= \left( \Partial{F_z}{y} - \Partial{F_y}{z} \right) dy\wedge dz + \left( \Partial{F_x}{z} - \Partial{F_z}{x} \right) dz\wedge dx \nonumber\\ &\qquad + \left( \Partial{F_y}{x} - \Partial{F_x}{y} \right) dx\wedge dy \end{align} But these components look just like those of $\grad\times\FF$, except that we have taken a 1-form to a 2-form! We can undo this using the Hodge dual. Thus, under the usual identification \begin{equation} F = \FF\cdot d\rr \end{equation} of vectors and 1-forms, we also have \begin{equation} dF = (\grad\times\FF) \cdot d\AA \end{equation} or equivalently \begin{equation} {*}dF = (\grad\times\FF) \cdot d\rr \end{equation}

Similarly, starting from a 2-form \begin{equation} G = F_x \,dy\wedge dz + F_y \,dz\wedge dx + F_z \,dx\wedge dy \end{equation} we ask what $dG$ is, and compute \begin{align} dG &= dF_x\wedge dy\wedge dz + dF_y\wedge dz\wedge dx + dF_z\wedge dx\wedge dy \nonumber\\ &= \left(\Partial{F_x}{x}+\Partial{F_y}{y}+\Partial{F_z}{z}\right) dx\wedge dy\wedge dz \end{align} This looks like $\grad\cdot\FF$, except that it's a 3-form, namely \begin{equation} dG = (\grad\cdot\FF)\,dV \end{equation} Furthermore, $G$ is just ${*}F$. Putting this all together, we see that \begin{equation} {*}d{*}F = \grad\cdot\FF \end{equation}

Finally, returning to the gradient, we can similarly write \begin{equation} df = \grad f \cdot d\rr \end{equation} Thus, as suggested in §Differentiation of Differential Forms, each of the vector calculus operators div, grad, and curl has an analog in the language of differential forms. Explicitly, if $f$ is a function and $F$ is the 1-form corresponding to the vector field $\FF$, then $df$ is the 1-form corresponding to $\grad f$, ${*}dF$ is the 1-form corresponding to $\grad\times\FF$, and ${*}d{*}F$ is the function $\grad\cdot\FF$. Furthermore, each of these operators is really just the exterior derivative $d$, acting as \begin{equation} \bigwedge\nolimits^0 \longrightarrow \bigwedge\nolimits^1 \longrightarrow \bigwedge\nolimits^2 \longrightarrow \bigwedge\nolimits^3 \end{equation}