Exercise 2.08

Verify (2.78): for the exterior derivative of a product of a p-form $\omega$ and a q-form $\eta$, we have $$ d(\omega \land\eta) =(d\omega)\land \eta + (-1)^{p}\;\omega \land (d \eta ) $$

$$ \begin{align} \\ d(\omega \land\eta) &= (p+q+1)\partial_{[\mu_{1}}(\omega \land \eta)_{\mu_{2}\dots\mu_{p+q+1}]} \\ &=(p+q+1)\frac{(p+q)!}{p!\,q!}\partial_{[\mu_{1}}\omega_{[\mu_{2}\dots\mu_{p+1}}\eta_{\mu_{p+2}\dots\mu_{p+q+1}]]} \\ &=\frac{(p+q+1)!}{p!\,q!}\partial_{[\mu_{1}}\omega_{\mu_{2}\dots\mu_{p+1}}\eta_{\mu_{p+2}\dots\mu_{p+q+1}]} \\ &=\frac{(p+q+1)!}{p!\,q!}\Big(\eta_{[\mu_{p+2}\dots\mu_{p+q+1}}\partial_{\mu_{1}}\omega_{\mu_{2}\dots\mu_{p+1}]}+(-1)^{p}\omega_{[\mu_{2}\dots\mu_{p+1}}\partial_{\mu_{1}}\eta_{\mu_{p+2}\dots\mu_{p+q+1}]} \Big) \\ \end{align} $$
$$ \begin{align} (d\omega)\land \eta &=(p+1)\partial_{[\mu_{1}}\omega_{\mu_{2}\dots \mu_{p+1}]}\land \eta_{\mu_{p+2}\dots \mu_{p+q+1}} \\ &=(p+1)\frac{(p+q+1)!}{(p+1)!\,q!}\eta_{[\mu_{p+2}\dots \mu_{p+q+1}}\partial_{[\mu_{1}}\omega_{\mu_{2}\dots \mu_{p+1}]]} \\ &=\frac{(p+q+1)!}{p!\,q!}\eta_{[\mu_{p+2}\dots \mu_{p+q+1}}\partial_{\mu_{1}}\omega_{\mu_{2}\dots \mu_{p+1}]} \end{align} $$
$$ \begin{align} \omega \land (d\eta) &=\omega_{\mu_{2}\dots \mu_{p+1}}\land(q+1)\partial_{[\mu_{1}} \eta_{\mu_{p+2}\dots \mu_{p+q+1}]} \\ &=(q+1) \frac{(p+q+1)!}{p!\,(q+1)!}\omega_{[\mu_{2}\dots \mu_{p+1}}\partial_{[\mu_{1}} \eta_{\mu_{p+2}\dots \mu_{p+q+1}]]} \\ &=\frac{(p+q+1)!}{p!\,q!}\omega_{[\mu_{2}\dots \mu_{p+1}}\partial_{\mu_{1}} \eta_{\mu_{p+2}\dots \mu_{p+q+1}]} \end{align} $$
So we indeed find the relation $(2.78)$. $$ \boxed{ d(\omega \land\eta) =(d\omega)\land \eta + (-1)^{p}\;\omega \land (d \eta )} $$