Show $L_{x}(w\wedge n)=L_{x}w\wedge n+w\wedge L_{x}n$

Question

I am asked to show the identity: $L_{x}(w\wedge n)=L_{x}w\wedge n+w\wedge L_{x}n$, where $w$ is a $k$-form and $n$ is a $l$-form.

I have finished most part of the proof, but I was stuck in the end. Here is my partial proof:

Bt Cartan Formula, $L_{x}(w\wedge n)=[i(X)d+di(X)](w\wedge n)=i(X)d(w\wedge n)+di(X)(w\wedge n)$.

Since $d$ is an antiderivative of degree $=1$.

Then, we have the equation $=i(X)[dw\wedge n+(-1)^{k}w\wedge dn]+di(X)(w\wedge n)$.

Moreover, since $i(X)$ is the antiderivative of degree $=-1$.

Then, we have the equation $=i(X)[dw\wedge n+(-1)^{k}w\wedge dn]+d[i(X)w\wedge n+(-1)^{k}w\wedge i(X)n]=i(X)dw\wedge n+(-1)^{k}w\wedge i(X)dn+di(X)w\wedge n+(-1)^{k}w\wedge di(X)n=[i(X)d+di(X)](w)\wedge n+(-1)^{k}w\wedge [i(X)d+di(X)](n)$

Then, we don't know how to next, how can I get rid of the (-1)^{k}?

Thank you so much for any hints and explanations!

Answer 1

$=i(X)[dw\wedge n+(-1)^{k}w\wedge dn]+d[i(X)w\wedge n+(-1)^{k}w\wedge i(X)n]$

this is right now, then the formula can be expanded into eight terms:

$=i(X)dw\wedge n+(-1)^{k+1}dw\wedge i(X)(n)+(-1)^{k}i(X)(w)\wedge dn+(-1)^{2k}w\wedge i(X)dn\\ +d(i(X)w)\wedge n+(-1)^{k-1}i(X)w\wedge dn+(-1)^{k}dw\wedge i(X)n+(-1)^{2k}w\wedge d(i(X)n)$

four terms are elimited, and the left are

$=[i(X)d+di(X)](w)\wedge n+(-1)^{2k}w\wedge [i(X)d+di(X)](n)$