The exterior derivative is a graded derivation

In a chart, $df={\sum }_j({\mathrm{\partial }}_{j}f)d{x}^j$, and [F1] expresses $d\omega$ by differentiating each coefficient and adding one coordinate differential. Real linearity of partial differentiation therefore makes $d$ real linear on each degree, and every resulting term has degree one higher. Extending by the finite homogeneous decomposition gives a linear map on ${\mathrm{\Omega }}^{\ast }$.

Write $\alpha ={\sum }_I{a}_{I}d{x}^I$ and $\beta ={\sum }_J{b}_{J}d{x}^J$. The ordinary coefficient product rule gives $d(a_I{b}_J)=b_{J}d{a}_I+a_{I}d{b}_J$. Thus [F1] applied termwise to their wedge product gives $d\alpha \wedge \beta$ from the first summands. In the second summands, [F3] gives $d{b}_J\wedge d{x}^I=(-1{)}^{p}d{x}^I\wedge d{b}_J$, yielding $(-1{)}^p\alpha \wedge d\beta$. Repeated coordinate indices give zero wedges on both sides.

By [F2], these chart identities are the restrictions of the corresponding global forms; equality on a chart cover implies equality on $M$. This globalizes linearity, the degree shift and the product rule, which together are exactly [F4]. The calculation includes degree zero via the empty wedge, and degrees above the dimension give zero.