text: Q:
  
  Some calculations with skew forms and wedge product
  
  I have some problems with the language of multilinear forms.
  I have to prove that if $dim(V)\le 3$, then every $\omega\in\Lambda^q(V^\ast)$ is such that $\omega\wedge\omega=0$.
  I consider the case $n:=dim(V)=3$.
  I can write 
  $$
  \omega=\sum_{1\le i_1<\dots<i_q\le 3}\omega_{i_1\dots i_q}\epsilon_{i_1}\wedge\dots\wedge\epsilon_{i_q}=
  \sum_{1\le i_1<i_2<i_3\le 3}\omega_{i_1i_2i_3}\epsilon_{i_1}\wedge\epsilon_{i_2}\wedge\epsilon_{i_3}
  $$
  Then I make a lot of confusions because of the notation and I can't go on with the calculations. Any help?
  
  A:
  
  As by the question linked by Michael, it is generally proved in a course about differential forms that $\alpha\wedge\beta = (-1)^{pq} \beta\wedge\alpha$ where $p$ and $q$ are the degrees of $\alpha$, $\beta$. Using this the result is immediate.
  But you could try and prove it, at least in your specific case. Hint: it should not be difficult to prove it for basic 3-forms, that is 3-forms which are product of 1-forms, $\omega=\alpha\wedge\beta\wedge\gamma$, it is basically the definition of skew form. Now how does the general case follow from the one for basic 3-forms?
  EDIT
  I misread the question and my answer was partially incorrect.
  If the degree of the form is odd the result holds generally (that is no matter what the dimension of the vector space $n$ is) and can be proved along the lines suggested above. If the degree is even the result generally does not hold. In fact it does not hold for 0-forms which are just constants. For 2-forms it holds, quite trivially because the space of 4-forms on a vector space of dimension 3 has dimension 0.
  
  