Lie Derivative of one-form: an identity

[Quality Cheese]

Homework Statement

I am trying to prove an identity for the Lie derivative of a smooth one-form. The identity is: for X, Y smooth vector fields, alpha a smooth one-form, we have:
$$L_{[X, Y]}\alpha = [L_X, L_Y]\alpha$$ For anyone familiar with the book, this is exercise 5.26 in the first edition of Nakahara: Geometry, Topology, and Physics.

Homework Equations

I am given the identity: for X, Y smooth vector fields, alpha a smooth one-form,
$$(L_X\alpha)(Y)=L_X(\alpha(Y))-\alpha([X, Y])$$ ([X, Y] is the Lie Bracket of the vector fields X and Y, and $$L_XY=[X, Y])$$

The Attempt at a Solution

I keep trying to expand the lie derivatives and cancel terms but I think I am missing a property of the lie derivative; maybe I'm messing up the Leibniz rule or something? Any help would be greatly appreciated!

 

[Nino MMP]

A:By definition, $L_{[X,Y]}\alpha=d(i_{[X,Y]}\alpha)+i_{[X,Y]}d\alpha$. $L_X\alpha = d(i_X\alpha)+i_Xd\alpha$ and $L_Y\alpha = d(i_Y\alpha)+i_Yd\alpha$.From the leibniz rule:$$(L_X\circ L_Y-L_Y\circ L_X)\alpha = d(i_Xd(i_Y\alpha)-i_Yd(i_X\alpha))+i_Xd(i_Yd\alpha)-i_Yd(i_Xd\alpha))$$Now, using the identity you have been given,$$L_{[X,Y]}\alpha=d(i_{[X,Y]}\alpha)+i_{[X,Y]}d\alpha=d(i_Xd(i_Y\alpha)-i_Yd(i_X\alpha))+i_Xd(i_Yd\alpha)-i_Yd(i_Xd\alpha))=$$$$(L_X\circ L_Y-L_Y\circ L_X)\alpha$$