Let
\[\omega^p = \sum_{i_1 <...< i_p} f_{i_1} ...f_{i_p} dx_{i_1} \wedge ... \wedge dx_{i_p}\]
\[\omega^q = \sum_{j_1 <...< j_q} g_{j_1} ...g_{j_q} dx_{j_1} \wedge ... \wedge dx_{j_q}\]
The exterior product of
\[\omega^p\]
and \[\omega^q\]
i\[\omega^p \wedge \omega^q = \sum f_{{i_1} ...{i_p}} g_{{j_1} ...{j_q}} dx_{i_1} \wedge ... \wedge dx_{i_p} \wedge dx_{j_1} \wedge ... \wedge dx_{j_q} \]
(1)We can reverse the order of
\[\omega^p , \: \omega^q\]
to give\[\omega^p \wedge \omega^q =(-1)^{p+q} \omega^q \wedge \omega^p\]
Proof
The summation in (1) includes the factors
\[ dx_{i_1} \wedge ... \wedge dx_{i_p} \wedge dx_{j_1} \wedge ... \wedge dx_{j_q}\]
Interchanging consecutive
\[dx_k\]
changes the sign of the whole expression so moving \[dx_{j_1}\]
past the \[p\]
terms to the left introduces a factor \[(-1)^p\]
and we get\[ dx_{i_1} \wedge ... \wedge dx_{i_p} \wedge dx_{j_1} \wedge ... \wedge dx_{j_q}=(-1)^p dx_{j_1} \wedge dx_{i_1} \wedge ... \wedge dx_{i_p} \wedge dx_{j_2} \wedge ... \wedge dx_{j_q} \]
Moving each of the
\[dx_{j_k}\]
past the \[p\]
\[dx{i_m}\]
introduces a factor \[(-1)^p\]
and since there are \[q\]
\[dx_{j_m}\]
to move, we introduce a factor \[(-1)^{pq}\]
when this is finished hence \[\omega^p \wedge \omega^q =(-1)^{p+q} \omega^q \wedge \omega^q\]