Cartan commutator identities

Both $[{\mathcal{L}}_X,{\iota }_Y]$ and ${\iota }_{[X,Y]}$ are degree-$-1$ graded derivations, so it suffices to compare them on functions and one-forms. They vanish on functions. For a one-form $\alpha$, the tensor derivation rule gives $({\mathcal{L}}_X\alpha )(Y)=X(\alpha (Y))-\alpha ({\mathcal{L}}_{X}Y)=X(\alpha (Y))-\alpha ([X,Y])$. Hence $[{\mathcal{L}}_X,{\iota }_Y]\alpha ={\mathcal{L}}_X(\alpha (Y))-({\mathcal{L}}_X\alpha )(Y)=\alpha ([X,Y])={\iota }_{[X,Y]}\alpha$. The equality extends to all forms by the derivation rules.

The commutator $[{\mathcal{L}}_X,{\mathcal{L}}_Y]$ is a degree-zero derivation. On a function $f$ it is $X(Yf)-Y(Xf)=[X,Y]f$. On a one-form $\alpha$, evaluate the commutator at a vector field $Z$ and use the one-form formula from step 1.1 twice. The terms where $X$ and $Y$ differentiate $\alpha (Z)$ combine to $[X,Y](\alpha (Z))$; the remaining terms combine by the Jacobi identity to $-\alpha ([[X,Y],Z])$. This is $({\mathcal{L}}_{[X,Y]}\alpha )(Z)$, proving the second identity on all forms. Finally, for any alternating form $\omega$, $({\iota }_X{\iota }_Y\omega )(Z_1,\dots )=\omega (Y,X,Z_1,\dots )=-\omega (X,Y,Z_1,\dots )=-({\iota }_Y{\iota }_X\omega )(Z_1,\dots )$. This proves the third identity, including degrees zero and one.