Let $\omega$ be the $2$-form in $\mathbb{R}^{2n}$ given by $$\omega=dx_1 \wedge dx_2+dx_3\wedge dx_4+\cdots dx_{2n-1}\wedge dx_{2n}$$ Compute the exterior product of $n$ copies of $\omega$.
(Chapter 1, Exercise 7 in Differential Forms and Applications by Manfredo P. do Carmo)
The wording of the problem is kind of difficult for me to understand, but I suppose that the exercise asks us to calculate $\omega \wedge \cdots \wedge \omega$.
So, I think the only way one can get a non-zero coefficient is that if we permute the parentheses $(dx_1 \wedge dx_2)$ up to $(dx_{2n-1}\wedge dx_{2n})$. There are $n!$ such permutations. Since each permutation can be arranged to $dx_1\wedge dx_2 \wedge dx_3 \wedge dx_4 \wedge \cdots \wedge dx_{2n-1}\wedge dx_{2n}$ with an even number of transpositions, we always get $+1$ as the coefficient of the differential form. So, the answer should be
$$\omega\wedge\cdots\wedge\omega=\color{green}{n!}\,dx_1\wedge dx_2 \wedge dx_3\wedge dx_4 \wedge \cdots \wedge dx_{2n-1}\wedge dx_{2n}$$
Is that right? If I write my argument exactly like here, would it be considered a complete calculation and receive full points in an exam of differential manifolds?
Your argument is correct, except for the fact that you need to justify why each of your permuted pairs of $dx_{i}\wedge dx_{i+1}$ when wedged together form $dx_1\wedge\cdots\wedge dx_{2n}$ without a sign. This is because: given forms $\alpha,\beta$ of degrees $k,\ell$ respectively, $$\alpha\wedge\beta=(-1)^{k\ell} \beta\wedge \alpha. $$ In particular, here $\deg(\alpha)=\deg(\beta)=2$ so that $$ (dx_{i}\wedge dx_{i+1})\wedge (dx_j\wedge dx_{j+1})=(-1)^4 (dx_j\wedge dx_{j+1})\wedge (dx_i\wedge dx_{i+1})=(dx_j\wedge dx_{j+1})\wedge (dx_i\wedge dx_{i+1})$$ for any choice of $i\ne j$ (both even).