We now investigate the properties of exterior differentiation.

Consider $d^2f=d(df)$. We have \begin{align} d^2 f = d(df) &= d\left(\Partial{f}{x^i}\,dx^i\right) \nonumber\\ &= d\left(\Partial{f}{x^i}\right)\wedge dx^i \nonumber\\ &= \frac{\partial^2f}{\partial x^j\partial x^i} \, dx^j \wedge dx^i = 0 \end{align} since mixed partial derivatives are independent of the order of differentiation, and the wedge product is antisymmetric. More formally, interchanging the dummy indices $i$ and $j$ changes the sign in the wedge product, but not the coefficient, thus demonstrating that $d^2f$ is equal to its negative, and must therefore be zero.

A similar argument works for differential forms of arbitrary rank. Suppose that \begin{equation} \alpha = f \, dx^I \end{equation} so that \begin{equation} d\alpha = df \wedge dx^I \end{equation} Then \begin{align} d^2 \alpha = d(d\alpha) &= d\left(\Partial{f}{x^i}\,dx^i\wedge dx^I\right) \nonumber\\ &= d\left(\Partial{f}{x^i}\right)\wedge dx^i\wedge dx^I \nonumber\\ &= \frac{\partial^2f}{\partial x^j\partial x^i} \, dx^j\wedge dx^i\wedge dx^I = 0 \end{align} This result is often summarized by writing \begin{equation} d^2 = 0 \end{equation}

In three Euclidean dimensions, this rule reduces to well-known identities. We have \begin{align} \grad\times\grad f &\longleftrightarrow {*}d(df) = {*}ddf = 0 \\ \grad\cdot(\grad\times\FF) &\longleftrightarrow {*}d{*}({*}dF) = *d{*}{*}dF = *ddF = 0 \end{align} Thus, the familiar identities $\grad\times\grad f=0$ and $\grad\cdot(\grad\times\FF)=0$ are both special cases of $d^2=0$.

Like all differential operators, $d$ also satisfies a product rule. With $\alpha$ a $p$-form as above, suppose that \begin{equation} \beta = h \, dx^J \end{equation} is a $q$-form. Then \begin{align} d(\alpha\wedge\beta) &= d \left( (f\,dx^I) \wedge (h\,dx^J) \right) \nonumber\\ &= d \left( fh\, dx^I \wedge dx^J \right) \nonumber\\ &= d(fh) \wedge dx^I \wedge dx^J \nonumber\\ &= (h\,df+f\,dh) \wedge dx^I \wedge dx^J \nonumber\\ &= h\,df \wedge dx^I \wedge dx^J + f\,dh \wedge dx^I \wedge dx^J \nonumber\\ &= (df \wedge dx^I) \wedge h\,dx^J + (-1)^p f\,dx^I\wedge (dh\wedge dx^J) \nonumber\\ &= d\alpha \wedge \beta + (-1)^p \alpha \wedge d\beta \end{align} where the factor of $(-1)^p$ arises from moving $dh$ through the $p$-form $dx^I$.