We divide the proof in several steps.

Step One.

Suppose $M=(0,1]\times(0,1)^{{n-1}}$ and

(i.e. the term $dx_{j}$ is missing). Hence we have

and from the definition of integral on a manifold we get

From the fundamental theorem of Calculus we get

Since $\omega$ and hence $f$ have compact support in $M$ we obtain

On the other hand we notice that $\int_{{\partial M}}\omega$ is to be understood as $\int_{{\partial M}}i^{*}\omega$ where $i:\partial M\to M$ is the inclusion map. Hence it is trivial to verify that when $j\neq 1$ then $i^{*}\omega=0$ while if $j=1$ it holds

and hence, as wanted

Step Two.

Suppose now that $M=(0,1]\times(0,1)^{{n-1}}$ and let $\omega$ be any differential form. We can always write

and by the additivity of the integral we can reduce ourself to the previous case.

Step Three.

When $M=(0,1)^{n}$ we could follow the proof as in the first case and end up with $\int_{M}d\omega=0$ while, in fact, $\partial M=\emptyset$.

Step Four.

Consider now the general case.

First of all we consider an oriented atlas $(U_{i},\phi_{i})$ such that either $U_{i}$ is the cube $(0,1]\times(0,1)^{{n-1}}$ or $U_{i}=(0,1)^{n}$. This is always possible. In fact given any open set $U$ in $[0,+\infty)\times\mathbb{R}^{{n-1}}$ and a point $x\in U$ up to translations and rescaling it is possible to find a “cubic” neighbourhood of $x$ contained in $U$.

Then consider a partition of unity $\alpha_{i}$ for this atlas.

From the properties of the integral on manifolds we have

The second integral in the last equality is zero since $\sum_{i}d\alpha_{i}=d\sum_{i}\alpha_{i}=0$, while applying the previous steps to the first integral we have

On the other hand, being $(\partial U_{i},\phi_{{|\partial U_{i}}})$ an oriented atlas for $\partial M$ and being ${\alpha_{i}}_{{|\partial U_{i}}}$ a partition of unity, we have

and the theorem is proved.