Let $n \geqslant 2$ be an integer and consider the vector space $V=F^{2n}$ with the standard basis $e\_1,\ldots,e\_{2n}$.

Now the second exterior power $\Lambda^2(V)$ contains the element $$\omega=e\_1 \wedge e\_2 + e\_3 \wedge e\_4 + \cdots + e\_{2n-1} \wedge e\_{2n}.$$

Since $\Lambda^{2n}(V)$ is spanned by $x := e\_1 \wedge \cdots \wedge e\_{2n}$, we know that the $n$-fold wedge product $\omega \wedge \cdots \wedge \omega$ is a scalar multiple of $x$. 

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> Question: How do we compute this scalar?
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