Q: Computing a scalar $c$ such that $cx= c(e_1 \wedge \cdots \wedge e_{2n})=\omega \wedge \cdots \wedge \omega$. 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 $x$. 

Question: How do we compute this scalar?


A: First distribute all terms in the product defining $\omega \wedge \cdots \wedge \omega$ (there are $n$ products). Notice that all of the terms with repeated $x_j$ will vanish by antisymmetricity of the wedge product. Thus you get $$ \omega^{\wedge n} = \sum_{\sigma \in S^*_n} e_{\sigma(1)}\wedge e_{\sigma(1)+1} \wedge \cdots \wedge e_{\sigma(n)}\wedge e_{\sigma(n)+1}$$
where $S_n^*$ denotes the group of all bijections from $\{1,3,5,...,2n-1\}$ unto itself. Next notice that if $\sigma \in S_n^*$, then $e_{\sigma(1)}\wedge e_{\sigma(1)+1} \wedge \cdots \wedge e_{\sigma(n)}\wedge e_{\sigma(n)+1} = e_1 \wedge \cdots \wedge e_n$, because reordering the terms two at a time does not change the sign of the $2n$-form. Consequently, $$\omega^{\wedge n} = \big| S_n^* \big| \cdot e_1 \wedge \cdots \wedge e_n = n! \cdot e_1 \wedge \cdots \wedge e_n$$

A: Here is a somewhat more general method that may complement shalop's answer. Let $A_{ij}$ be a matrix of dimension $2n$ and $\widetilde{A}_{ij} = \frac{A_{ij}-A_{ji}}{2}$ be the antisymmetric part. Then let $\omega = \sum_{i,j=1}^{2n} A_{ij} e_i \wedge e_j = \sum_{i,j=1}^{2n} \widetilde{A}_{ij} e_i \wedge e_j$.
Then it is easy to show that $\omega \wedge \dots \wedge \omega = 2^n n! \mathrm{pf}(\widetilde{A}) e_1 \wedge \dots \wedge e_{2n}$ where pf means Pfaffian.
The Pfaffian has the property that $\mathrm{pf}(\widetilde{A})^2 = \det(\widetilde{A})$. In your case $\widetilde{A} =  \left( \begin{matrix} 0 & \frac{1}{2} \\ -\frac{1}{2} & 0 \end{matrix}\right) \oplus \dots \oplus \left( \begin{matrix} 0 & \frac{1}{2} \\ -\frac{1}{2} & 0 \end{matrix}\right)$. From this you have $\mathrm{pf}(\widetilde{A}) = \frac{1}{2^n}$ from which it follows that in your case that $c = 2^n n! \mathrm{pf}(\widetilde{A}) = n!$.
