Proving $[L_X,i_Y]=[i_X,L_Y]=i_{[X,Y]}$

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: $$[i_X, \mathcal{L}_Y] = -[\mathcal{L}_Y, i_X] = -i_{[Y, X]} = i_{-[Y, X]} = i_{[X, Y]}.$$