Q:

Show that $d(\omega ∧ \tau ) = (d\omega)∧\tau +(−1)^{\text{deg}\omega} w ∧d\tau.$

The exterior differentiation $d : \Omega^∗(U)→\Omega^∗(U)$ is an
  antiderivation of degree $1$: $$d(\omega ∧ \tau ) = (d\omega)∧\tau
 +(−1)^{\text{deg}\omega} w ∧d\tau.$$ 

Suppose $\omega=\sum a_Idx^I$ and $\tau=\sum b_Jdx^J$
$d(\omega ∧ \tau )=d(\sum_{I\cap J=\phi} a_Ib_Jdx^I\wedge dx^J).$
$=\sum_{I\cap J=\phi} d(a_Ib_J)\wedge (dx^I\wedge dx^J)$
$=\sum_{I\cap J$=\phi} \sum_{k=1}^n\frac{\partial(a_Ib_J)}{\partial x^k}dx^k \wedge (dx^I\wedge dx^J) $ 
$=\sum_{I\cap J=\phi} \sum_{k=1}^n(a_I\frac{\partial(b_J)}{\partial x^k}+\frac{\partial(a_I)}{\partial x^k}b_J)dx^k \wedge (dx^I\wedge dx^J) $
$=\sum_{I\cap J=\phi} \sum_{k=1}^n(a_I\frac{\partial(b_J)}{\partial x^k}+\frac{\partial(a_I)}{\partial x^k}b_J)dx^k \wedge dx^I\wedge dx^J$
$=\sum_{I\cap J=\phi} \sum_{k=1}^n a_I\frac{\partial(b_J)}{\partial x^k}dx^k \wedge dx^I\wedge dx^J+ \sum_{I\cap J=\phi} \sum_{k=1}^n\frac{\partial(a_I)}{\partial x^k}b_J dx^k \wedge dx^I\wedge dx^J$
........................................................................................................................................................
$= \sum_{k=1}^n \frac{\partial(b_J)}{\partial x^k}dx^k\sum_{I\cap J=\phi}a_I \wedge dx^I\wedge dx^J+ \sum_{k=1}^n\frac{\partial(a_I)}{\partial x^k}dx^k \sum_{I\cap J=\phi}b_J   \wedge dx^I\wedge dx^J$
Am I correct? how do I complete the proof? Please help me.

A:

You are correct.
To complete the proof, in the first summand of your last line, permute $dx^{I}$ to the left so that it is adjacent to $a_{I}$.
In the second summand, move $b_{J}$ to the right so that it is adjacent to $dx^{J}$.

