Prove the formula for the Lie derivative of a differential form

Question

If $X$ is a vector field then by $\mathcal F^t_X$ I will denote it's flow. If $\alpha \in \Lambda^k$ then by definition $$ \mathcal L_X \alpha = \frac{d}{dt}(\mathcal F^t_X)^*\alpha \, \biggr|_{t=0}. \tag{1} $$ The following formula holds: $$ \mathcal L_X \alpha = d(\alpha \rfloor X) +(d\alpha)\rfloor X. \tag{2} $$ M. Taylor at page 78 of "Partial Differential Equations I" writes that (2) implies the formula $$ (\mathcal L_X \alpha)(X_1,\ldots,X_k) = X \cdot \alpha(X_1,\ldots,X_k)-\sum_j \alpha(X_1,\ldots,[X,X_j],\ldots,X_k), $$ if we take into account that $\mathcal L_X X_j = [X,X_j]$. But how to see this? I tried to write $$ \bigl((\mathcal F^t_X)^* \alpha \bigr)(X_1,\ldots,X_k) = \alpha(\mathcal F^t_{X\#} X_1,\ldots,\mathcal F^t_{X\#}X_k)\circ \mathcal F^t_X $$ and then differentiate it, taking into account that partial derivative of a multilinear function is the function itself, but I don't see how this approach uses the formula (2). Please, help me.

Answer 1

HINT: If $\alpha$ is a $k$-form, find the formula for $d\alpha(X,X_1,\dots,X_k)$. When $k=1$, the famous formula is $$d\alpha(X,Y) = X(\omega(Y))-Y(\omega(X))-\omega([X,Y]).$$