Let $\omega$ be a 2-differential form in $\mathbb{R}^{2n}$ given by
$$\displaystyle \omega=dx^1\wedge dx^2+dx^3\wedge dx^4 + \cdots + dx^{2n-1}\wedge dx^{2n}$$
Compute:
$$\displaystyle \overbrace{\omega\wedge\omega\wedge\cdots\wedge\omega}^{\text{n times}}.$$
Then I have already done the case n=1,2,3 so I have in general that
1) $\omega\wedge\omega$ is 4-form
2) $\omega\wedge\omega\wedge\omega$ is a 6 -form
$\vdots$
n-1) ($\omega\wedge\omega\wedge\cdots\wedge\omega$) is a 2n-form
therefore we have the relation as in the n step we have a (2k+1)-form where k is the number of wedges that apear, so we have that product will be zero since we have all the 2n members of the basis ($dx^i$) in the for of the step n-1 so the product should be zero.
Am I right? or How can I do this in a better way?
Thanks a lot for your help in advance :).
I don't know where you're getting the $(2k+1)$-form. If you wedge $\omega$ with itself $k$ times, you get — as you yourself observed — a $2k$-form. Let's try an example: With $n=2$ and $\omega = dx_1\wedge dx_2+dx_3\wedge dx_4$, we'll have $\omega\wedge\omega = 2 dx_1\wedge dx_2\wedge dx_3\wedge dx_4$. Since $2$-forms commute with one another, you should be able to show that for the general case you get $$\underbrace{\omega\wedge\dots\wedge\omega}_{n\text{ times}} = n!\,dx_1\wedge\dots\wedge dx_{2n}.$$