An exact nonzero multiple satisfies the Pfaff condition

Fix $q\in U$ and distinct coordinates $i,j$. Choose $h,k>0$ small enough that the coordinate rectangle from $q$ to $q+h{e}_i+k{e}_j$ lies in $U$, and set $\mathrm{\Delta }=S(q+h{e}_i+k{e}_j)-S(q+h{e}_i)-S(q+k{e}_j)+S(q)$. By [F4] the second partials are uniformly continuous on the chosen closed rectangle. Applying [F2] to coordinate segments and [F3] to their continuous derivatives, Newton–Leibniz gives $\mathrm{\Delta }={\int }_0^h\text{\hspace{0.17em}⁣}{\int }_0^k{S}_{ij}(q+s{e}_i+t{e}_j)dtds={\int }_0^k\text{\hspace{0.17em}⁣}{\int }_0^h{S}_{ji}(q+s{e}_i+t{e}_j)dsdt\mathrm{.}$ Each equality follows by taking the two endpoint differences in the outer coordinate and then applying Newton–Leibniz to the inner difference; both integrands are continuous by $S\in C^2$.

Divide both expressions in step 1.1 by $hk$ and let $h,k\downarrow 0$. Continuity of $S_{ij}$ and $S_{ji}$ makes each average tend to its value at $q$, so $S_{ij}(q)=S_{ji}(q)$. Since $q,i,j$ were arbitrary, the coordinate formula in [F1] gives $d(dS)=0$.

Put $\lambda :=1\mathrm{/}\mu$. Total differentiability gives $\mu (q+h)=\mu (q)+D\mu (q)h+o(\mathrm{\parallel }h\mathrm{\parallel })$, and continuity with $\mu (q)\mathrm{\ne }0$ gives $\mu (q+h)\mu (q)=\mu (q{)}^2+o(1)$. Hence $\lambda (q+h)-\lambda (q)=-(\mu (q+h)-\mu (q))\mathrm{/}(\mu (q+h)\mu (q))=-\mu (q{)}^{-2}D\mu (q)h+o(\mathrm{\parallel }h\mathrm{\parallel })$, so $\lambda \in C^1(U)$ with continuous derivative $D\lambda =-{\mu }^{-2}D\mu$. For each $i$, $S_i={\mathrm{\partial }}_{i}S$ is $C^1$. The increment identity $\lambda (q+h)S_i(q+h)-\lambda (q)S_i(q)=\lambda (q+h)(S_i(q+h)-S_i(q))+S_i(q)(\lambda (q+h)-\lambda (q))$ and the total derivative expansions therefore give $D(\lambda S_i)=\lambda D{S}_i+S_{i}D\lambda$. Since $\omega =\lambda dS$, for coordinates $j<k$ the coefficient of $d{x}_j\wedge d{x}_k$ in $d\omega$ is ${\mathrm{\partial }}_j(\lambda S_k)-{\mathrm{\partial }}_k(\lambda S_j)=({\mathrm{\partial }}_j\lambda )S_k-({\mathrm{\partial }}_k\lambda )S_j$, as the mixed partials cancel by step 2.1. Thus $d\omega =d\lambda \wedge dS$, and $\omega \wedge d\omega =\lambda dS\wedge d\lambda \wedge dS=0$ because the alternating product vanishes when $dS$ occurs twice.