= Solution
Let $e$ be the identity. For a <left-invariant differential form> $\omega$ and a <left-invariant vector field> $X$, one has
$$
\omega_g(X_g)=\omega_g((dL_g)_eX_e)=(L_g^*\omega)_e(X_e)=\omega_e(X_e).
$$
The value is therefore independent of $g$, even if the <Lie group> is disconnected. The same calculation applies to another left-invariant field $Y$.
The standard <exterior derivative of a one-form evaluated on vector fields> is
$$
d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y]).
$$
Both differentiated functions are constant in the present situation. Thus
$$
\boxed{d\omega(X,Y)=-\omega([X,Y]).}
$$
This identity is a direct consequence of left invariance and the stated general formula for the <exterior derivative>.
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