Prove the identity $[L_X,L_Y]\omega=L_{[X,Y]}\omega$ for the Lie derivative of a $k$-form $\omega$

Question

Let $X$ and $Y$ be vector fields on $\mathbb{R}^n$.

Show that for $\omega$, a $k$-form on $\mathbb{R}^n$, $(L_XL_Y-L_YL_X)\omega=L_{[X,Y]}\omega $.

I try using Cartan's magic formula and get that

$\begin{align} (L_XL_Y-L_YL_X)\omega &= (L_XL_Y)\omega-(L_YL_X)\omega \\ &= L_X(L_Y\omega)-L_Y(L_X\omega) \\ &= L_X(i_Yd\omega+d(i_Y\omega))-L_Y(i_X(d\omega)+d(i_X\omega)) \\ &= i_Xdi_Yd\omega+i_Xd^2(\cdot) + di_Xi_Yd\omega+di_Xdi_Y\omega \\ &- i_Ydi_Xd\omega-i_Yd^2(\cdot) - di_Yi_Xd\omega-di_Ydi_X\omega \\ &= i_Xdi_Yd\omega + di_Xi_Yd\omega+di_Xdi_Y\omega - i_Ydi_Xd\omega - di_Yi_Xd\omega-di_Ydi_X\omega \\ &= [i_Xdi_Y-i_Ydi_X]d\omega+d[i_Xi_Y-i_Yi_X]d\omega+d[i_Xdi_Y-i_Ydi_X]\omega \end{align}$

For the other part I get

$L_{[X,Y]}\omega=i_{[X,Y]}d\omega+di_{[X,Y]}\omega$

From here I am now stuck.

For the first part of the question (which I have not posted) you are asked (and I am able to) show that $(L_XL_Y-L_YL_X)f=L_{[X,Y]}f$. Not sure whether this is relevant to my question about $\omega$.

Answer 1

Now apply the magic formula again: $$ L_X(\eta)=i_Xd\eta+di_X\eta $$ where $\eta=i_Yd\omega+di_Y\omega$ and get $$ L_X(L_Y\omega)=i_Xd(i_Yd\omega+di_Y\omega)+di_X(i_Yd\omega+di_Y\omega) $$ use that $d$ is linear $$ i_Xd(i_Yd\omega)+i_X(di_Y\omega)+di_X(i_Yd\omega)+di_X(di_Y\omega) $$ Subtract $L_Y L_X\omega$ and compare with $$ L_{[X,Y]}\omega=i_{[X,Y]}d\omega++di_{[X,Y]}\omega $$ (you should prove:) $$ [L_X,i_Y]=[i_X,L_Y]=i_{[X,Y]} $$