A good, intuitive explanation of $d(d\omega)=0$ for a differential form $\omega$? I was reading Munkres' Analysis on Manifolds and the book suggests that this is due to the fact that partial derivatives commute (i.e. $D_iD_jf=D_jD_if$ for functions $f\in C^\infty$), and some other posts suggests using the Stokes' Theorem ($\int_{M}d\omega=\int_{\partial M}\omega$).
I'd love to know if there is another more intuitive explanation. (i.e. a geometric one perhaps)
 A: This is not really an answer, but maybe just a little more to think about on the two aspects of $d$ which you mentioned:
Partials:
On the face of it, I've found that we seem to correlate "exterior derivative" and "partials commuting" as you mentioned. Since $dx^{i_1}\wedge \dotsm\wedge dx^{i_k}$ is anti-symmetric, it will "pick out" the alternating part of its coefficients. So
$$\sum a_{ij} dx^i\wedge dx^j = \sum \frac12(a_{ij}-a_{ji}) dx^i \wedge dx^j$$
and in the case of $d^2$ you can see it is $0$ iff the partial derivatives commute.
Later on you might see that if $\nabla$ is a covariant derivative operator (if you don't know what this is, just treat it like a special partial derivative operator, sort of like $\frac\partial{\partial x}$), then some its curvature $2$-form $F^{\nabla}$ can be defined as the failure of the second derivative to commute
$$F^{\nabla}_{X,Y} = \nabla_X\nabla_Y - \nabla_Y\nabla_X -\nabla_{[X,Y]}.$$
(The $[X,Y]$ term is included to make $F^\nabla$ tensorial.) Now, it is also possible to define an "exterior covariant derivative" operator $d^\nabla$, which extends the notion of your exterior $d$. Then you define $F^\nabla = (d^\nabla)^2$, so again $(d^\nabla)^2$ measures the failure of derivatives to commute.
Stokes:
There is another definition of $d$ which I remember from Vlad Arnold's (classical mechanics) book, which basically says that the operator $d$ is defined exactly to make Stokes' theorem true. He gives a pretty visual description of what $d\omega\in \Omega^k$, (something like "the $k$-linear part of $\int_{\partial P} \omega$ when you change the size of $P$, where $P$ is a small $k$-parallelpiped in your space"). So that might be something to look at.
