Stokes theorem implies that $\partial^2=0$? First of all, is this true?  Can you use Stokes' theorem plus $d^2 = 0$, or some other relatively easy to establish fact to prove that the boundary of a boundary is empty? If so, how exactly does one do this?
I've seen the argument (in Rudin p. 276, for example, see above) that, since $\int_{\partial\partial\Omega} \omega =\int_{\partial\Omega} d\omega = \int_\Omega d^2\omega$ for any smooth (or at least twice continuously differentiable) form $\omega$, and we know that $d^2\omega=0$, we get $\int_{\partial\partial\Omega}\omega = 0$. Since this is true for any $\omega$, that implies that $\partial\partial\Omega=\emptyset$.  
Why is that true?  Couldn't $\partial\partial\Omega$ be some nonempty but "small" domain such that you wind up integrating over a set of measure zero whenever you evaluate $\int_{\partial\partial\Omega}\omega$?
EDIT: I should clarify that I originally saw this in the context of integration over chains.  I am not certain how this relates to integration over manifolds. I guess these are different questions (too much of a noob...).  I don't see, for instance, how $\partial^2c=0$ for any $n$-chain $c$ would imply that $\partial^2\Omega=0$ for $n$-dimensional manifolds.
 A: The fact that $\partial\partial\Omega=\emptyset$ is trivial from the definition of the boundary of a smooth manifold and using Stokes' theorem is ridiculous overkill.  In this context, $\Omega$ is a smooth $n$-manifold with boundary, meaning it has coordinate charts that make it locally diffeomorphic to $[0,\infty)\times\mathbb{R}^{n-1}$ or $\mathbb{R}^n$ at each point.  The boundary of $\partial\Omega$ is defined as the union of the images of $\{0\}\times\mathbb{R}^{n-1}$ under the charts with domain $[0,\infty)\times\mathbb{R}^{n-1}$, and identifying $\{0\}\times\mathbb{R}^{n-1}$ with $\mathbb{R}^{n-1}$ these become the charts of an $(n-1)$-manifold structure on $\partial\Omega$.  Since all of these charts have domain $\mathbb{R}^{n-1}$, not  $[0,\infty)\times\mathbb{R}^{n-2}$, the boundary of $\partial\Omega$ is empty by definition.
That said, the Stokes' theorem argument does work.  If $M$ is an oriented $m$-manifold and $\int_M \omega=0$ for all smooth $m$-forms $\omega$ on $M$, then $M$ must be empty.  Indeed, if $M$ is nonempty, you can just use a bump function in any coordinate chart (times $dx_1\wedge dx_2\dots\wedge dx_m$ where the $x_i$ are the coordinate functions) to get an $m$-form whose integral is nonzero.  You don't have to worry about $M=\partial\partial\Omega$ being some weird "small domain" as you suggest since $\partial\partial\Omega$ is by definition a manifold.
