Question about the proof of Stokes' theorem. I'm trying to understand the proof of Stokes' theorem:

Given a compactly supported $m-1$-form $\omega$ on $\mathbb{R}^m$ then
$$\int_{\mathbb{R}^m} d\omega=0.$$

In the proof given by the teacher in his notes, he writes $\omega$ as 
\begin{equation}\omega=\frac{1}{m-1}\varepsilon_{i_1i_2\ldots i_m}f^{i_1}dx^{i_2}\wedge\ldots \wedge dx^{i_m},\;\;\;\;\;\text{(1)}\end{equation}
where $f^i$'s are compactly supported functions and $\varepsilon$ is the Levi-Civita symbol. He then claims that 
$$d\omega=\left(\frac{\partial f^1}{\partial x_1}+\cdots+\frac{\partial f^m}{\partial x^m}\right)dx^1\wedge\ldots\wedge dx^m.\;\;\;\;\;(2)$$
From the last equation one can easily deduce the theorem. What I don't undertand is how he wrote and arbitrary $m-1$-form $\omega$ as in (1) and then carry out the computation in (2).
 A: Every $(m-1)$-form is a linear combination of the forms
$$dx^1\wedge\dots\wedge dx^{m-1}, \quad dx^1\wedge\dots\wedge dx^{m-2}\wedge dx^m, \quad \dots \quad dx^1\wedge\dots\wedge dx^{i-1}\wedge dx^{i+1}\wedge\dots\wedge dx^m, \quad \dots \quad dx^2\wedge\dots\wedge dx^m,$$
as $i$ varies from $i$ to $m$ (i.e., you omit precisely one $dx^i$ and wedge the remaining basis $1$-forms).
Your professor is introducing lot of redundancy in his expression by allowing all multiindices $i_1i_2\dots i_m$. In fact, the $1/(m-1)$ should, I believe, be a factor of $1/(m-1)!$, to allow for all the permutations of $i_2,\dots,i_m$. The $\pm$ coming from the $\varepsilon$ symbol gets the signs right so that you have
\begin{align*}
d\big(\varepsilon_{i_1i_2\dots i_m} f^{i_1}\,dx^{i_2}\wedge\dots\wedge dx^{i_m}\big) &= \frac{\partial f^{i_1}}{\partial x^{i_1}}(\varepsilon_{i_1i_2\dots i_m}dx^{i_1}\wedge dx^{i_2}\wedge\dots dx^{i_m} )\\ &= \frac{\partial f^{i_1}}{\partial x^{i_1}}dx^1\wedge\dots\wedge dx^m.
\end{align*}
EDIT: The usual argument (with no $\varepsilon$ symbols needed) is this: Using my comment in the first paragraph, we write
$$\omega = \sum_{i=1}^m (-1)^{i-1} f^i dx^1\wedge\dots\wedge dx^{i-1}\wedge dx^{i+1}\wedge\dots\wedge dx^m.$$
Then the minus sign is cooked up just to make the following equation hold (after all, you must switch $dx^i$ with $dx^1\wedge\dots\wedge dx^{i-1}$ and this comes at a cost of $(-1)^{i-1}$):
$$d\omega = \sum_{i=1}^m \frac{\partial f^i}{\partial x^i} dx^1\wedge dx^2\wedge\dots\wedge dx^m.$$
