Bi-invariant form on compact connected Lie group

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Let $G$ be a Lie group and $\omega$ be a left invariant $k$-form, how to prove that $r^*_a \omega$ is left invariant? What I do:

$(l^*_g (r^*_a \omega))_x (v_1, \ldots,v_k)=(r^*_a \omega))_{gx} ({l_g}_*v_1, \ldots,{l_g}_*v_k)= \omega_{gxa} ({r_a}_* {l_g}_*v_1, \ldots,{r_a}_* {l_g}_*v_k)=$

and stuck here...

The right side of the equality should be $(r^*_a \omega)_x(v_1, \ldots,v_k)$.