Wedge product (Differential Geometry) How can we show $$\alpha_p\wedge\beta _q=(-1)^{pq}\beta_q\wedge\alpha_p$$ for the wedge product of a p- and q- form
 A: I assume you're familiar with the anti-commutative relation of the wedge product, $dx_i\wedge dx_j = -dx_j\wedge dx_i$. The idea is that we move terms over one-by-one, which adds powers to the minus sign.
It suffices to show this property for basic $p$ and $q$ forms with one term, since we can extend the argument linearly using using the properties of the wedge product. Let $\alpha_p = a_I dx_I$ and $\beta_q = b_J dx_J$. This notation just means that for $I = \{i_1, \dotsc, i_p\}$,
$$dx_I = dx_{i_1}\wedge\dotsm\wedge dx_{i_p},$$
and similarly for $J = \{j_1,\dotsc, j_q\}$. We then have the following computation:
\begin{align*}
\alpha_p\wedge\beta_q &=  a_Ib_J dx_I\wedge dx_J \\
& =  a_I b_J dx_{i_1}\wedge\dotsm\wedge dx_{i_p}\wedge dx_{j_1}\wedge\dotsm \wedge dx_{j_q} \\
& =  a_I b_J (-1)^p dx_{j_1} \wedge dx_{i_1}\wedge\dotsm\wedge dx_{i_p} \wedge dx_{j_2}\wedge\dotsm \wedge dx_{j_q} \\
& = a_Ib_I (-1)^{pq} dx_{j_1}\wedge \dotsm \wedge dx_{j_q}\wedge dx_{i_1}\wedge \dotsm \wedge dx_{i_p} = (-1)^{pq} \beta_q\wedge \alpha_p
\end{align*}
where we moved the $dx_{j_1}$ term $p$ positions over in the third line (giving us a factor of $(-1)^p$), and in the last line, we have shifted each of the remaining $(q-1)$ $dx_j$ forms $p$ positions to the left in the wedge product.
