The left-invariant Maurer-Cartan forms are given by $$g^{-1}dg, $$ where $g$ is a Lie group $G$ to $M_n(\mathbb{R})$.
My question is why is $$d(g^{-1}dg)=(g^{-1}dg)\wedge(g^{-1}dg)\quad ? $$ How come one gets an exterior product out of an exterior derivative?
$g^{-1}g=1$, so $dg^{-1}g+g^{-1}dg=0$, hence $dg^{-1}=-g^{-1}dg\cdot g^{-1}$. Thus $d(g^{-1}dg)=d(g^{-1})\wedge dg=-g^{-1}dg\cdot g^{-1}\wedge dg=g^{-1}dg\wedge g^{-1}dg$