The Exterior Derivative

Given a p - form  
  we can take the exterior derivative  
\[d \omega^p\]
  resulting in a p+1 - form.
  is a function. If  
\[\omega^0 =x^2_1 x_2 + x_2 x^2_3 \]
\[\begin{equation} \begin{aligned} d \omega^0 &= d(x^2_1 x_2 + x_2 x^2_3)\\ &=2x_1 x_2 dx_1 + x^2_1 dx_2 +x^2_3 dx_2 + 2x_2 x_3 dx_3 \\ &=2x_1 x_2dx_1 +(x^2_1+x^2_3)dx_2 + 2x_2 x_3 dx_3 \end{aligned} \end{equation}\]

\[\omega^1 =x^2_2 dx_1 +x_1 x_2 dx_2\]

\[\begin{equation} \begin{aligned} d (\omega^1 )& = d(x^2_2 dx_1 +x_1 x_2 dx_2) \\ &= =d(x^2_2) \wedge dx_1 + d(x_1 x_2) \wedge dx_2 \\ &= 2x_2 dx_2 \wedge dx_1 +x_2 dx_1 \wedge dx_2 + x_1 dx_2 \wedge dx_2 \\ &= 2x_2 dx_2 \wedge dx_1 +x_2 dx_1 \wedge dx_2 \\ &=-2x_2 dx_1 \wedge dx_2 +x_2 dx_1 \wedge dx_2 \\ &= - x_2 dx_1 \wedge dx_2 \end{aligned} \end{equation}\]

Sionce interchanging consecutive  
  changes the sign of a term.
\[\omega^2 =x_1 x_2 dx_1 \wedge dx_2 + x_2 x_3 dx_2 \wedge dx_3\]

\[\begin{equation} \begin{aligned} d \omega^2 &=d(x_1 x_2) \wedge dx_1 \wedge dx_2 + d(x_2 x_3) \wedge dx_2 \wedge dx_3 \\ &= x_2 dx_1 \wedge dx_1 \wedge dx_2 + x_1 dx_2 \wedge dx_1 \wedge dx_2 +x_3 dx_2 \wedge dx_2 \wedge dx_3 + x_2 dx_3 \wedge dx_2 \wedge dx_3 \\ &=0 \end{aligned} \end{equation} \]

since since any repeat of  
  in a term results in that term being zero.