The properties of $d$ in fact determine it uniquely.

Theorem: There is a unique operator $d:\bigwedge^p\longmapsto\bigwedge^{p+1}$ satisfying the following properties:

1. $d(a\,\alpha+\beta) = a\,d\alpha+d\beta$ if $a=\hbox{constant}$;
2. $d(\alpha\wedge\beta) = d\alpha\wedge\beta + (-1)^p \,\alpha\wedge d\beta$ if $\alpha$ is a $p$-form;
3. $d^2\alpha = 0$;
4. $df = \Partial{f}{x^i}\,dx^i$.

The first property is standard linearity, the second is the product rule; all differential operators satisfy a linearity condition and a product rule. The last condition ensures that $d$ does the right thing on functions, and could have been written \begin{equation} d(f) = df \end{equation}

Existence is straightforward: Properties 1 and 4 are obvious, and the remaining two properties were established in the previous section.

To show uniqueness, we have to demonstrate that the defining equation for $d$ follows from these properties. But the product rule with $p=0$ takes the form \begin{equation} d(f\,\alpha) = df\wedge\alpha + f\,d\alpha \end{equation} so that in particular \begin{equation} d(f\,dx^I) = df\wedge dx^I + f\,d\left(dx^I\right) \end{equation} and property 4 ensures that the first term is what we are looking for; it only remains to show that the last term is zero. But $d(dx^i)=0$ by property 3, after which $d(dx^i\wedge dx^j)=0$ follows by property 2. Continuing by induction, $d(dx^I)=0$ for any $I$, and we are done.