Given the $q$-form $\omega$, the $r$-form $\eta$, and the $s$-form $\xi$, show that the exterior product satisfies the properties:

(i) $\omega \wedge \omega=0$, when the order of $\omega$ is odd,

(ii) $\omega \wedge \eta=(-1)^{q r} \eta \wedge \omega$,

(iii) $(\omega \wedge \eta) \wedge \xi=\omega \wedge(\eta \wedge \xi)$.

For the purposes of this problem, it is possible to use the following expansions $$\begin{align*}
\omega & =\frac{1}{q !} \omega_{\mu_{1} \mu_{2} \ldots \mu_{q}} d x^{\mu_{1}} \wedge d x^{\mu_{2}} \wedge \cdots \wedge d x^{\mu_{q}} \tag{4.264}\\
\eta & =\frac{1}{r !} \eta_{v_{1} v_{2} \ldots v_{r}} d x^{v_{1}} \wedge d x^{v_{2}} \wedge \cdots \wedge d x^{v_{r}} \tag{4.265}\\
\xi & =\frac{1}{s !} \eta_{ho_{1} ho_{2} \ldots ho_{s}} d x^{ho_{1}} \wedge d x^{ho_{2}} \wedge \cdots \wedge d x^{ho_{s}} \tag{4.266}
\end{align*}$$
and direct computations.