aka de Rham differential

On any manifold $X$, there is a natural linear map $d: \Omega^k(X) \to \Omega^{k+1}(X)$ (where $\Omega^k(X)$ is the space of k-forms on manifold $X$) called the exterior derivative or the de Rham differential satisfying

1. If $f \in \Omega^0(X) = C^\infty(X)$ then $df \in \Omega^1 (X) = \Gamma^\infty(T^*X)$ is the usual Derivative of $f$
2. $d^2=0: \Omega^k(X) \to \Omega^{k+1}(X)$
3. d(\alpha \wedge \beta) = (d\alpha)\wedge \beta+(-1)^k \alpha \wedge(d \beta) if $\alpha \in \Omega^k(X)$, $\beta\in \Omega^l(X)$.

These properties characterize $d$ uniquely.

Geometrical intuition behind this?

In coordinates:

$d\alpha_U = \sum\limits_{j=1,...,n \: ; \: i_1...i_k: 1\leq i_1<...<i_k \leq n} \frac{\partial\alpha_{i_1...i_k}}{\partial x_j}dx_j \wedge dx_{i_1} \wedge...\wedge dx_{i_k}$

where $\alpha_{i_1...i_k}$ are the components of the Exterior form $\alpha$. Note that the new exterior form has indices which may not be ordered in increasing order. So it would be easier if $\alpha$ was in that form too...