Proof of Brouwer fixed point theorem using Stokes's theorem $\omega$ is the volume form on the boundary
B -ball
$f\colon B \to \partial B$
$$
0 < \int_{\partial B}\omega = \int_{\partial B} f^*(\omega) = \int_{B} df^*(\omega) = \int_B f^*(d\omega) = \int_B(0) = 0
$$
What I do not understand, is why on the left, we may conclude that


*

*$\int_{B}  d \omega = \int_{\partial B}  \omega > 0  $


where as later we are alowed to conclude


*$\int_{B}  f^*(d\omega$) =0


as $d\omega=0$
What puzzles me is, what prevents us from concluding that $d\omega=0$ in 1. 
Both 1. and 2. are integrated over B, so where does the difference come from?

In the equation
$$
    \int_{\partial \Omega} \alpha = \int_{\Omega} d\alpha
$$
it is assumed that $\alpha$ is defined on $\Omega$.  If $i\colon\partial\Omega\to\Omega$ is the inclusion, it induces a restriction map $i^*$ on differential forms in the opposite direction. Really the $\alpha$ on the left-hand side is $i^*\alpha$. We don't usually bother with this extra notation, but it matters.
In general $i^*$ is not an injection.  Let's consider the $n=2$ case.  So $B$ is the unit disk and $\partial B$ is the unit circle.  Let $\alpha = x\,dy - y\,dx$ and $\omega = i^*\alpha$.  In other words, $\alpha$ is an extension of $\omega$ from $\partial B$ to $B$.  We have
$$
    d\alpha = 2\, dx\wedge dy
$$
which is definitely not zero on the unit disk.
But $i^*(d\alpha)=di^*\alpha = d\omega$ is zero, by dimension count (it's a two-form on a one-manifold).
To say the same thing without the $i^*$, it is possible for $\omega$ to extend from $\partial B$ to $B$, and for $d\omega =0$ to be true on $\partial B$, but $d\omega \neq 0$ on $B$.  This is what's happening here.
Now back to your proof.  Suppose $\omega$ is a volume form on $\partial B$ (we are no longer assuming that $\omega = i^*\alpha$ for some $\alpha$ on $B$; maybe it is, maybe it isn't).  If $f \colon B \to \partial B$ were a retraction, then $f \mathbin{\circ} i = \operatorname{id}_{\partial B}$.  Then on differential forms, $i^* \mathbin{\circ} f^* = \operatorname{id}_{\Omega^*(\partial B)}$.  Therefore
$$\begin{split}
\int_{\partial B}\omega &= \int_{\partial B}i^*(f^*\omega) \\
&= \int_B d(f^*\omega) \qquad\text{(by Stokes)} \\
&= \int_B f^*(d\omega) \\
&= \int_B f^*(0) \qquad\text{(by dimension count)} \\
&= 0
\end{split}$$
If $\omega$ is a volume form on $\partial B$, this is a contradiction.
