Proof of Formua for Interchanging Two Forms

\[\omega^p = \sum_{i_1 <...< i_p} f_{i_1} ...f_{i_p} dx_{i_1} \wedge ... \wedge dx_{i_p}\]

\[\omega^q = \sum_{j_1 <...< j_q} g_{j_1} ...g_{j_q} dx_{j_1} \wedge ... \wedge dx_{j_q}\]

The exterior product of  
\[\omega^p \wedge \omega^q = \sum f_{{i_1} ...{i_p}} g_{{j_1} ...{j_q}} dx_{i_1} \wedge ... \wedge dx_{i_p} \wedge dx_{j_1} \wedge ... \wedge dx_{j_q} \]
We can reverse the order of  
\[\omega^p , \: \omega^q\]
  to give
\[\omega^p \wedge \omega^q =(-1)^{p+q} \omega^q \wedge \omega^p\]

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  
  changes the sign of the whole expression so moving  
  past the  
  terms to the left introduces a factor  
  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  
  past the  
  introduces a factor  
  and since there are  
  to move, we introduce a factor  
  when this is finished hence  
\[\omega^p \wedge \omega^q =(-1)^{p+q} \omega^q \wedge \omega^q\]