MHB Why is No One Answering This Week's Problem?

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Here's this week's problem!

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Problem. Let $\nu$ be the differential 2-form $dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}$ on $\Bbb R^{2n}$. Show that $\nu^n = n!\, dx_1 \wedge dx_2 \wedge dx_3 \wedge \cdots \wedge dx_{2n}$.
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No one answered this week's problem. You can find my solution below.

Spoiler

Let $\nu_j := dx_{2j-1} \wedge dx_{2j}$, for $j = 1,2,\cdots, n$. Then the nonzero contributions to $\nu^n$ are terms of the form $\nu_J$ for some ascending index $J$ of order $n$. For each ascending index $J$, there corresponds a unique permutation $\pi \in S_n$ such that $\nu_J = \nu_{\pi(1)} \wedge \cdots \wedge \nu_{\pi(n)}$. So $\nu^n$ is the sum of terms $\nu_{\pi(1)}\wedge \cdots \wedge \nu_{\pi(n)}$, as $\pi$ ranges over $S_n$. Since each $\nu_j$ is a two-from, $\nu_j \wedge \nu_k = (-1)^{2\cdot 2} \nu_k \wedge \nu_j = \nu_k \wedge \nu_j$ for all $k \neq j$. It follows that $\nu_{\pi(1)}\wedge \cdots \wedge \nu_{\pi(n)} = \nu_1 \wedge \cdots \wedge \nu_n$ for all $\pi \in S_n$. Thus

$$\nu^n = \sum_{\pi\in S_n} \nu_1 \wedge \cdots \wedge \nu_n = \#(S_n)\, \nu_1 \wedge \cdots \wedge \nu_n = n!\, dx_1 \wedge \cdots \wedge dx_{2n}.$$