# Let $w_1, \ldots, w_n$ be left invariant $1$-forms on a Lie group $G$. Why is $w = w_1 \wedge \ldots \wedge w_n$ left-invariant?

$\ldots$while the $w_i$ are dual to vector fields $X_1 , \ldots , X_n$ spanning $T_e G$. I cannot apply the definition straight forward because I do not really no how to handle the wedge product in combination with the left invariance.

In general $F^*(\alpha\wedge \beta) = (F^*\alpha)\wedge(F^*\beta)$ so
$$L_g^*(w_1\wedge\dots\wedge w_n) = (L_g^*w_1)\wedge\dots\wedge(L_g^*w_n) = w_1\wedge\dots\wedge w_n.$$
Note, the fact that $\{w_i\}$ are dual to $\{X_i\}$ was not used in the above computation.
If $T : V \to W$ is a linear operator on a vector space $V$, then the induced map $\wedge^k(T) : \wedge^k(V) \to \wedge^k(W)$ acts by $T (v_1 \wedge \ldots \wedge v_n) = (Tv_1) \wedge \ldots \wedge (Tv_n)$.