Proving $[L_X,i_Y]=[i_X,L_Y]=i_{[X,Y]}$ Let $X,Y$ be vector fields. $L_X$ is the Lie derivative and $i_X$ is the contraction of a $k$-form. 
I am really stuck on how you could prove the identity $[L_X,i_Y]=[i_X,L_Y]=i_{[X,Y]}$.
Update: I have the definition $L_X \omega = \frac{\partial}{\partial t}|_{t=0} \Phi_t^* \omega$, where $\Phi_t$ is the flow of $\mathbb{X}$.
I have not seen the formula $(\mathcal{L}_X\alpha)(V_1, \dots, V_k) = X(\alpha(V_1, \dots, V_k)) - \sum_{i=1}^k\alpha(V_1, \dots, V_{i-1}, [X, V_i], V_{i+1}, \dots, V_k)$.
 A: These identities can be found using the formulae $(i_Y\alpha)(V_1, \dots, V_k) = \alpha(Y, V_1, \dots, V_k)$ and 
$$(\mathcal{L}_X\alpha)(V_1, \dots, V_k) = X(\alpha(V_1, \dots, V_k)) - \sum_{i=1}^k\alpha(V_1, \dots, V_{i-1}, [X, V_i], V_{i+1}, \dots, V_k).$$
First of all, we have
\begin{align*}
& (\mathcal{L}_Xi_Y\alpha)(V_1, \dots, V_k)\\
=&\ (\mathcal{L}_X(i_Y\alpha))(V_1, \dots, V_k)\\
=&\ X((i_Y\alpha)(V_1, \dots, V_k)) - \sum_{i=1}^k(i_Y\alpha)(V_1, \dots, V_{i-1}, [X, V_i], V_{i+1}, \dots, V_k)\\
=&\ X(\alpha(Y, V_1, \dots, V_k)) - \sum_{i=1}^k\alpha(Y, V_1, \dots, V_{i-1}, [X, V_i], V_{i+1}, \dots, V_k).
\end{align*}
Now note that
\begin{align*}
& (i_Y\mathcal{L}_X\alpha)(V_1, \dots, V_k)\\
=&\ (i_Y(\mathcal{L}_X\alpha))(V_1, \dots, V_k)\\
=&\ (\mathcal{L}_X\alpha)(Y, V_1, \dots, V_k)\\
=&\ X(\alpha(Y, V_1, \dots, V_k)) - \alpha([X, Y], V_1, \dots, V_k)\\ 
&\ - \sum_{i=1}^k\alpha(Y, V_1, \dots, V_{i-1}, [X, V_i], V_{i+1}, \dots, V_k)\\
=&\ (\mathcal{L}_Xi_Y\alpha)(V_1, \dots, V_k) - \alpha([X, Y], V_1, \dots, V_k)\\
=&\ (\mathcal{L}_Xi_Y\alpha)(V_1, \dots, V_k) - (i_{[X, Y]}\alpha)(V_1, \dots, V_k).
\end{align*}
After rearranging we conclude that 
$$[\mathcal{L}_X, i_Y] = \mathcal{L}_Xi_Y - i_Y\mathcal{L}_X = i_{[X, Y]}.$$ 
One can do a similar calculation to prove the other identity. Alternatively, we can deduce it directly from the previous result as follows: $$[i_X, \mathcal{L}_Y] = -[\mathcal{L}_Y, i_X] = -i_{[Y, X]} = i_{-[Y, X]} = i_{[X, Y]}.$$
