We refer to the product of two 1-forms as a 2-form. The space of all 2-forms, denoted $\bigwedge^2(\RR^n)$ or simply $\bigwedge^2$, is therefore spanned by all products of basis 1-forms, that is \begin{equation} \bigwedge\nolimits^2 = \left\langle\{dx^i \wedge dx^j\}\right\rangle \end{equation} But this set is redundant; it is sufficient to assume $i<j$. Similarly, we can define $p$-forms as \begin{equation} \bigwedge\nolimits^p = \left\langle\{dx^{i_1} \wedge … \wedge dx^{i_p}\}\right\rangle \end{equation} where $1\le i_1 < … < i_p\le n$.

Each of the spaces $\bigwedge^p$ is itself a vector space (or module), and since we have an explicit basis, we can determine its dimension, which is \begin{equation} \dim\left(\bigwedge\nolimits^p\right) = {n\choose p} = \frac{n!}{p!(n-p)!} \end{equation} at least so long as $p\le n$. What if $p=0$? We define $\bigwedge^0$ to be the space of scalars (functions), with dimension $1$. What if $p>n$? Since there are only $n$ independent 1-forms, any product with more than $n$ factors must be zero. Thus, \begin{equation} \dim\left(\bigwedge\nolimits^p\right) = 0 \qquad\qquad (p>n) \end{equation}