Wedge product computation Let $\omega \in \Omega^{2}(\mathbb{R}^{2n})$ be the $2$-form $\omega=dx^{1} \wedge dx^{2} + dx^{3} \wedge dx^{4} + \dots + dx^{2n-1} \wedge dx^{2n}$.
I want to compute the wedge product of $\omega$ with itself $n$ times. 
I know that $\omega \in \Omega^{2}(\mathbb{R}^{2n})$ so $\omega \wedge\dots\wedge \omega \in \Omega^{2+\dots+2}(\mathbb{R}^{2n}) = \Omega^{2n}(\mathbb{R}^{2n})$. Also the basis of $\Omega^{2n}(\mathbb{R}^{2n})$ is $dx^{1} \wedge dx^{2} \wedge dx^{3} \wedge dx^{4} \wedge \dots \wedge dx^{2n-1} \wedge dx^{2n}$. 
I computed for $n=1$, $n=2$, and $n=3$, obtaining $\omega=dx^{1} \wedge dx^{2}$, 
$\omega \wedge \omega = 2 dx^{1} \wedge dx^{2} \wedge dx^{3} \wedge dx^{4}$ and $\omega \wedge \omega \wedge \omega= 6 dx^{1} \wedge dx^{2} \wedge  dx^{3} \wedge dx^{4} \wedge dx^{5} \wedge dx^{6}$ respectively. 
How can I get the coefficient which multiplies the basis for the general $n^{\text{th}}$ wedge product case? 
Thanks.
 A: Let $\omega_i = dx^{2i-1}\wedge dx^{2i}$, so $\omega = \omega_1 + \dots + \omega_n$. We have 
$$\omega^n = \underbrace{(\omega_1 + \dots + \omega_n)\wedge\dots\wedge(\omega_1 + \dots + \omega_n)}_{n\ \text{copies of}\ \omega}.$$
Every term in the expansion takes the form $\omega_{i_1}\wedge\dots\wedge\omega_{i_n}$ where $i_1, \dots, i_n \in \{1, \dots, n\}$. In total, there are $n^n$ terms in the expansion, but many of these are zero. When is $\omega_{i_1}\wedge\dots\wedge\omega_{i_n} \neq 0$? Precisely when $i_1, \dots, i_n$ are distinct, so $i_1, \dots, i_n$ is a permutation of $1, \dots, n$. As every such permutation occurs in the expansion, we see that 
$$\omega^n = \sum_{\sigma \in S_n}\omega_{\sigma(1)}\wedge\dots\wedge\omega_{\sigma(n)}.$$
As $\omega_i$ is a two-form, $\omega_i\wedge\omega_j = \omega_j\wedge\omega_i$; in particular, $\omega_{\sigma(1)}\wedge\dots\wedge\omega_{\sigma(n)} = \omega_1\wedge\dots\wedge\omega_n$. Therefore
$$\omega^n = \sum_{\sigma \in S_n}\omega_1\wedge\dots\wedge\omega_n = |S_n|\,\omega_1\wedge\dots\wedge\omega_n = n!\,dx^1\wedge dx^2\wedge\dots\wedge dx^{2n-1}\wedge dx^{2n}.$$
A: As you know, for every $i$, $dx^i\wedge dx^i=0$, and so after opening the brackets, the terms that survive are only those where each $dx^i$ appears exactly once. Computing the number of these terms is a not-too-hard exercise in combinatorics.
