I have proved the following statement and I would like to have some feedback on my proof (is it correct? can it be improved?). Thanks.
Write the coordinates on $\mathbb{R}^{2n}$ as $(x_1,y_1,x_2,y_2,\dots, x_n, y_n).$ Let $\omega=dx_1dy_1+dx_2dy_2+\dots+dx_ndy_n=\sum\limits_{i=1}^{n}dx_idy_i.$ Compute $\omega^n=\omega\omega\cdots\omega$ ($n$-fold product).
My proof:
For $n=1,\ n=2,\ n=3,\ n=4$ we obtain, after multiplying all the terms and rearranging, respectively:
$\omega^1=dx_1dy_1, \omega^2=(dx_1dy_1+dx_2dy_2)(dx_1dy_1+dx_2dy_2)=2dx_1dy_1dx_2dy_2$ $\omega^3=(dx_1dy_1+dx_2dy_2+dx_3dy_3)(dx_1dy_1+dx_2dy_2+dx_3dy_3)(dx_1dy_1+dx_2dy_2+dx_3dy_3)=6dx_1dy_1dx_2dy_2dx_3dy_3$ $\omega^4=(dx_1dy_1+dx_2dy_2+dx_3dy_3+dx_4dy_4)(dx_1dy_1+dx_2dy_2+dx_3dy_3+dx_4dy_4)(dx_1dy_1+dx_2dy_2+dx_3dy_3+dx_4dy_4)(dx_1dy_1+dx_2dy_2+dx_3dy_3+dx_4dy_4)=24dx_1dy_1dx_2dy_2dx_3dy_3dx_4dy_4.$
Looking at these cases we make the hypothesis that $\omega^n=n!dx_1dy_1\cdots dx_ndy_n$ corroborated by the idea that in $\omega^n$ we have $n$ parentheses and after we choose a term $dx_idy_i$ from the first one, that term, due to the anticommutativity of the differential forms, can only give non-zero terms to the result after multiplying $n-1$ terms in the second parenthesis (all except the same $dx_idy_i$ term), so there are $n\cdot (n-1)$ such terms, and these new terms $dx_idy_idx_jdy_j$, for the same reason, can only multiply the terms in the third parenthesis that do not contain neither $dx_idy_i$ and $dx_jdy_j$ and there are $n-2$ such terms so we are now at a total of $n\cdot (n-1)\cdot (n-2)$ non-zero terms. Note that each multiplication gives an additional term since $dx_1dy_1\cdots dx_idy_i\cdots dx_jdy_j\cdots dx_ndy_n$ with $1\leq j<i\leq n$ can be rearranged to be in increasing order in an even number of switches because if there are $m$ pairs $dx_idy_k$ between $dx_idy_i$ and $dx_jdy_j$ then both $dx_j$ and $dy_j$ have to switch $2m$ times to be in increasing order hence $dx_1dy_1\cdots dx_idy_i\cdots dx_jdy_j\cdots dx_ndy_n=(-1)^{2m}dx_1dy_1\cdots dx_jdy_j\cdots dx_idy_i\cdots dx_ndy_n=dx_1dy_1\cdots dx_jdy_j\cdots dx_idy_i\cdots dx_ndy_n$. Continuining in this fashion we see that the result will be $n!dx_1dy_1dx_2dy_2\cdots dx_{n-1}dy_{n-1}dx_ndy_n.$
We prove this rigorously by induction. The case $n=1$ has already been shown above. Suppose now the result holds for $n\in N, n>1$: then by denoting $\omega_0:=dx_1dy_1+\dots +dx_ndy_n$ we get
\begin{align*} \omega^{n+1}&=\underbrace{\omega\cdots\omega}_{n\text{ times}}\omega=\underbrace{(\omega_0+dx_{n+1}dy_{n+1})\cdots (\omega_0+dx_{n+1}dy_{n+1})}_{n\text{ times}}(\omega_0+dx_{n+1}dy_{n+1})\\ &=(\omega_0+dx_{n+1}dy_{n+1})^{n+1}=\sum\limits_{k=0}^{n}\binom{n+1}{k}\omega_0^k(dx_{n+1}dy_{n+1})^{n+1-k}\\ &=\binom{n+1}{0}\omega_0^0(dx_{n+1}dy_{n+1})^{n+1}+\binom{n+1}{1}\omega_0^1(dx_{n+1}dy_{n+1})^n+\binom{n+1}{2}\omega_0^2(dx_{n+1}dy_{n+1})^{n-2}\\ &+\dots+\binom{n+1}{k}\omega_0^k(dx_{n+1}dy_{n+1})^{n-k}+\dots+\binom{n+1}{n-2}\omega_0^{n-2}(dx_{n+1}dy_{n+1})^3\\ &+\binom{n+1}{n-1}\omega_0^{n-1}(dx_{n+1}dy_{n+1})^2+\binom{n+1}{n}\omega_0^n(dx_{n+1}dy_{n+1})^1+\binom{n+1}{n+1}\omega_0^{n+1}(dx_{n+1}dy_{n+1})^0\\ &=\frac{(n+1)!}{n!1!}\omega_0^n(dx_{n+1}dy_{n+1})+\omega_0^{n+1}=(n+1)\omega_0^n(dx_{n+1}dy_{n+1})+\omega_0^n\omega_0\\ &\overset{ind.hyp.}{=} (n+1)(n!dx_1dy_1\cdots dx_ndy_n)(dx_{n+1}dy_{n+1})+(n!dx_1dy_1\cdots dx_ndy_n)(dx_1dy_1+\dots +dx_ndy_n)\\ &=(n+1)n!dx_1dy_1\cdots dx_ndy_ndx_{n+1}dy_{n+1}+0\\ &= (n+1)!dx_1dy_1\cdots dx_ndy_ndx_{n+1}dy_{n+1} . \end{align*} and this concludes the inductive step and thus the proof. $\square$
Your proof seems correct. I feel however that proceding by induction is not really necessary, as you can appeal to combinatorics.
Let me explain better what I mean: to ease notation, let $\eta_i=\mathrm{d}x_i\wedge\mathrm{d}y_i$, so that $\omega$ is the sum of the $2$-forms $\eta_i$. As you correctly noted, $\eta_i\wedge\eta_i=0$ and $\eta_i\wedge\eta_j=\eta_j\wedge\eta_i$, so we can already say that $\omega^n$ will be a multiple of $\eta_1\wedge\ldots\wedge\eta_n$.
To determine what the multiple will be, notice that to compute $\omega^n$ we are choosing one element of $\eta_1,\dots,\eta_n$ from each of the $n$ copies of $\omega$, and if two choices are equal the result is zero. Hence you are just counting the number of possible permutations of $n$ objects. In other words, \begin{equation} \omega^n=\left(\sum_{i=1}^n\eta_i\right)^n=\sum_{\sigma\in S_n}\eta_{\sigma(1)}\wedge\ldots\wedge\eta_{\sigma(n)}=\sum_{\sigma\in S_n}\eta_1\wedge\ldots\wedge\eta_n. \end{equation} Hence $\omega^n$ equals $\eta_1\wedge\ldots\wedge\eta_n$ multiplied by the cardinality of the symmetric group of $n$ elements.