Let $ \omega = dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n} \in \mathbb{R}^{2n}$. Find $\omega^n$ (in respect to $\wedge$) Let $ \omega = dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n} \in \mathbb{R}^{2n}$. Find $\omega^{n}$ (in respect to $\wedge$)
When I say "$\omega^{n}$ (in respect to $\wedge$)" I mean the exterior product of $n$ $\omega$'s.
I tried writing
$$(dx_1 \wedge dx_2) (v_1,v_2) = \det \begin{pmatrix}
d{x_1}(v_1) & d{x_1}(v_2)\\ 
d{x_2}(v_1) & d{x_2}(v_2)
\end{pmatrix} = d{x_1}(v_1)d{x_2}(v_2) - d{x_1}(v_2)d{x_2}(v_1)
$$
$$
(dx_3 \wedge dx_4) (v_1,v_2) = \det \begin{pmatrix}
d{x_3}(v_1) & d{x_3}(v_2)\\ 
d{x_4}(v_1) & d{x_4}(v_2)
\end{pmatrix} = d{x_3}(v_1)d{x_4}(v_2) - d{x_3}(v_2)d{x_4}(v_1)
$$
but that wouldn't lead me nowhere.
I know the property $ \omega \wedge (\phi_1 + \phi_2) = \omega \wedge \phi_1 + \omega \wedge \phi_2 $
Then:
$$
\begin{align}
\omega \wedge \omega = {} & (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \\
& {} \wedge (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \\[8pt]
= {} & (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \wedge (dx_1 \wedge dx_2) \\
& {} + (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \wedge (dx_3 \wedge dx_4) + \cdots \\
& {} + (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \wedge (dx_{2n-1} \wedge dx_{2n})
\end{align}
$$ but I'm not sure that's what I'm supposed to do...
Thanks.
 A: Write $\omega = \alpha_1 + \dots + \alpha_n$ where $\alpha_i = dx_{2i - 1} \wedge dx_{2i}$. The important observation is that if you expand
$$ \omega^n = (\alpha_1 + \dots + \alpha_n) \wedge \dots \wedge (\alpha_1 + \dots + \alpha_n) = \sum_{i_1, \dots, i_n} \alpha_{i_1} \wedge \dots \wedge \alpha_{i_n} = \sum_{I} \alpha_I$$
then if $I$ contains a repeated index the $2n$-form $\alpha_I$ will be a wedge of $2n$ one-forms in which (at least) two factors will be the same and so $\alpha_I = 0$. Thus,
$$ \omega^n = \sum_{\sigma \in S_n} \alpha_{\sigma(1)} \wedge \dots \wedge \alpha_{\sigma(n)}. $$
Since $\alpha_i$ are two-forms, we have $\alpha_{\sigma(1)} \wedge \dots \wedge \alpha_{\sigma(n)} = \alpha_1 \wedge \dots \wedge \alpha_n$ and so
$$ \omega^n = n! ( \alpha_1 \wedge \dots \wedge \alpha_n) = n! (dx_1 \wedge dx_2 \wedge \dots \wedge dx_{2n-1} \wedge dx_{2n}). $$
To demonstrate the argument for $n = 2$, we have
$$ \omega^2 = (dx_1 \wedge dx_2 + dx_3 \wedge dx_4) \wedge (dx_1 \wedge dx_2 + dx_3 \wedge dx_4) = \\
(dx_1 \wedge dx_2) \wedge (dx_1 \wedge dx_2) + (dx_3 \wedge dx_4) \wedge (dx_3 \wedge dx_4) + \\
(dx_1 \wedge dx_2) \wedge (dx_3 \wedge dx_4) + (dx_3 \wedge dx_4) \wedge (dx_1 \wedge dx_2)   = \\
2(dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4).$$
