The exterior derivative is the unique (sequence of) linear map $d: \mathcal{A}^p (M) \to \mathcal{A}^{p+1}$, such that the following axioms hold:

- for a function $f$, $df$ is the total differential.
- For any function $f$ and any differential form $a$, the Leibniz rule $d(fa)= df \wedge a + f da$ holds.
- For any diffeomorphism $\phi: M \to N$, you have $\phi^{\ast} \circ d = d \circ \phi^{\ast}$.

I think that 3 is more natural or at least easier to motivate than the usual $dd=0$. But both properties are really equivalent.

Proof (of uniqueness): 2. implies locality, i.e. the value of $d a$ at a point $x \in M$ only depends on the value of $a$ in a neighborhood of $x$. This, together with the axiom 3, shows that it is enough to consider $M =\mathbb{R}^n$.

The group $\mathbb{R}^n$ acts by translations on $\mathbb{R}^n$. By axiom 3, for any translation-invariant form $a$ on $\mathbb{R}^n$, the form $da$ is again translation-invariant.

On the other hand, each nonzero $\lambda \in \mathbb{R}$ gives rise to the diffeomorphism $h_{\lambda}:x \mapsto \lambda x$ of $\mathbb{R}^n$. It is easy to check that it acts on translation-invariant $p$-forms by multiplication with $\lambda^p$. Thus for any translation-invariant $p$-form $a$, you get

$$\lambda^p d a = d (\lambda^p a) = d (h_{\lambda}^{\ast} a ) = h_{\lambda}^{\ast} d a = \lambda^{p+1} da,$$

which implies that any translation-invariant form is closed. Finally, note that any $p$-form on $\mathbb{R}^n$ can be written as a linear combination of translation-invariant form, with coefficients in $C^{\infty}(\mathbb{R}^n)$ (a basis for the translation-invariant forms is formed by the usual elements $dx_{i_1} \wedge \ldots \wedge x_{i_p}$).

From axioms 1 and 2, you now conclude that $d$ must be the exterior derivative that you knew before. This, of course, implies all the other properties of $d$.