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There isn't much you need to do. The wedge product satisfies the relationship: $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ if $\alpha, \beta \in \Lambda^p, \Lambda^q$ respectively. In your case $$\omega \wedge \omega = (-1)^{(2q+1)(2q+1)} \omega \wedge \omega = -\omega \wedge \omega$$ That can only happen if $\omega \wedge \omega = 0$.
Comments to the question (v2): One can define left and right exterior derivatives $$d_L(\omega\wedge\eta)~=~(d_L\omega)\wedge\eta + (-1)^{|\omega|}\omega\wedge d_L\eta \tag{L},$$ $$d_R(\omega\wedge\eta)~=~(-1)^{|\eta|} (d_R\omega)\wedge\eta + \omega\wedge d_R\eta \tag{R},$$ $$d_R\omega~=~(-1)^{|\omega|}d_L \omega, \tag{C}$$ where $\omega, ... 1 Given a$p$-form$\theta \in \bigwedge^p E^*$, we can define an alternating multilinear map$h \colon E^{n-p} \to \bigwedge^n E$by $$h(u_1, \ldots, u_{n-p}) = \theta \wedge \tilde{u}_1 \wedge \ldots \wedge \tilde{u}_{n-p}.$$ Let$b \colon \mathbb{R} \to \bigwedge^n E$be the linear map $$b(t) = t \omega.$$ Because$\bigwedge^n E$is one-dimensional and ... 1 So let's assume that$V$has a non-degenerate bilinear form$\langle\cdot,\cdot\rangle$with a basis$e_1,\dots,e_n$such that$\langle e_i,e_j\rangle = \delta_{ij}$, the Kronecker delta. Let$*$denote the Hodge star operator. Note that we have the formula $$\langle x,y\rangle = *((*x)\wedge y) .$$ Let's identify any operator on$V$with its matrix ... 1 As by the question linked by Michael, it is generally proved in a course about differential forms that$\alpha\wedge\beta = (-1)^{pq} \beta\wedge\alpha$where$p$and$q$are the degrees of$\alpha$,$\beta\$. Using this the result is immediate. But you could try and prove it, at least in your specific case. Hint: it should not be difficult to prove it for ...