When talking about the wedge (exterior) product with regard to real differential $k$-forms, I know that if $\alpha$ is a $k_1$-form and $\beta$ is a $k_2$-form, then their wedge product is a $(k_{1}+k_{2})$-form such that $$\alpha \wedge \beta = (-1)^{k_{1}k_{2}}\beta \wedge \alpha\text{.}$$ In particular, if $k =k_1 = k_2= 1$ and $\alpha = \beta$, then $$\alpha \wedge \alpha =0\text {.} $$ This says that the wedge product of a $1$-form $\alpha$ with itself is $0$. But then what would be the wedge product of three such $1$-forms $\alpha$, i.e. $\alpha \wedge \alpha \wedge \alpha$? Would it also be $0$?
I'm certainly aware that the wedge product is associative. So clearly $$\begin{align*} (\alpha \wedge \alpha) \wedge \alpha &= \alpha \wedge (\alpha \wedge \alpha) \\ 0 \wedge \alpha &= \alpha \wedge 0 \end{align*} \text {,}$$ where $\alpha$ is a $1$-form. Now what is $0 \wedge \alpha$? Is $0$ a $0$-form?