I'm not sure if this point of view is taken up in the many references which are named here, but I'll say something about an "elementary" way to discover the exterior derivative which sounds like ordinary calculus.  Let's take on the point of view that a $k$-form is something you integrate over a $k$-dimensional submanifold.  If you imagine $k$-dimensional submanifolds as being composed of a $k$-dimensional blanket of $k$-parallelograms, then this is a geometrically natural point of view since the $k$-form will assign a number to each of these parallelograms.  To actually realize a submanifold as such a "blanket" is to give a parameterization.

In one-variable calculus, when $f$ is a function, $df$ tells you the change in $f$ per small change in its input, and if you integrate it over a curve from $a$ to $b$, it expresses the total change in $f$ from $a$ to $b$.  Now, a one form $\eta$ is integrated not over points but rather over curves.  Still, you can ask, how does $\int_\gamma\eta$ change when you perturb $\gamma$?  Well, if you deform a closed curve $\gamma_a$ into another curve $\gamma_b$, the difference between the integrals over $\gamma_b$ and $\gamma_a$ is some derivative we can call "$d\eta$" integrated over the surface.  

Picturing the case where $\gamma_a$ and $\gamma_b$ bound an annulus is a good thing to consider here; this interpretation tells you how to orient the boundary of the annulus if you want to think of $\int_\Sigma d\eta = \int_{\gamma_b} \eta - \int_{\gamma_a} \eta$ as being $\int_{\partial \Sigma} \eta$.  You can then contract the inner circle to a point (the integral over the inner circle will vanish by the linearity and continuity of the form) to recover Stokes' theorem for a disk.

It's not completely necessary that the curve (or $k$-dimensional submanifold) you deform is closed, but by rule the boundary should remain fixed during the deformation or you will miss out on part of the boundary.

Using a specific example like a square/cube, we can get a coordinate representation for $d\eta$ through the fundamental theorem of calculus.  (For $0$ forms, every point is closed, so we did not need to worry about the word "closed" before.)

It is easy to see many properties.  For example, let's take $\eta$ to be a $1$-form in $3$-space; then $d^2 \eta$ is clearly $0$.  Let $\gamma$ be a circle, and let $\Sigma_a$ and $\Sigma_b$ be the upper and lower hemispheres of a ball $B$ whose equator is $\gamma$.  Then $\int_{\Sigma_a} d \eta = \int_\gamma \eta = \int_{\Sigma_b} d \eta$ by Stokes' theorem for a disk.  On the other hand, the integral of $d^2\eta$ over the ball $B$ is just $\int_{\Sigma_b} d \eta - \int_{\Sigma_a} d\eta = 0$ because you can sweep out $B$ by deforming $\Sigma_a$ to $\Sigma_b$ with the boundary fixed.  Since $\int_B d^2 \eta = 0$ for every ball, $d^2 \eta$ is identically $0$.  When you execute this proof for a square, you see that mixed partials commute.

I would like to know if the product rule can easily be seen through this interpretation, but I have not thought enough about it to see it clearly yet.