Proof of Formua for Interchanging Two Forms

Theorem
Let
\[\omega^p = \sum_{i_1 <... i_p f_ ...f_ dx_ ...>
\[\omega^q = \sum_{j_1 <... j_q g_ ...g_ dx_ ...>
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  
\[\omega^p \wedge \omega^q =(-1)^{p+q} \omega^q \wedge \omega^q\]

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