I'm reading <Differential Geometry, Lie groups and Symmetric Spaces> by Helgason.
In Section 7 of Chapter 2, there is a lemma which says that the bi-invariant forms on a Lie group are closed. This question gives one way to prove it, and I also know the proof using the inversion map. However, the author go through with somewhat unclear method to me. He says:
$(\tilde{X}_i^{R(\exp tX)})_e = Ad(\exp (-tX))(X_i)$ for $X \in \mathfrak{g}$, and
$\sum_1^p \omega(\tilde{X}_1, \dots, [\tilde{X}, \tilde{X}_i], \tilde{X}_{i+1}, \dots, \tilde{X}_p) = 0$ by the derivative of $Ad$ and the right-invariance of $\omega$,
(Here, $\omega$ is a bi-invariant $p$-form and $\tilde{X}_i$ is a left-invariant vector field corresponding to the tangent vector $X_i \in T_e G$.)
and then using the exterior derivative formula for $d \omega$, he conclude that $d \omega = 0$. I can understand all the process, but above two formulae. How can I obtain them? Maybe they can be easily derived by definition of $Ad$ and the right-invariance, but I can't find a way now. Any help will be appreciated so much.