This is problem 20.8 from Loring Tu's An Introduction to Smooth Manifolds.
Bracket of the Lie Derivative and Interior Multiplication.
If $X$ and $Y$ are smooth vector fields on a manifold $M$, prove that on differential forms on $M$,
$$\mathcal{L}_X \iota_Y - \iota_Y \mathcal{L}_X = \iota_{[X,Y]}.$$
Hint: Let $\omega \in \Omega^k(M)$ and $Y, Y_1, \dots, Y_{k-1} \in \mathfrak{X}(M)$. Apply the global formula for $\mathcal{L}_X$ to $$(\iota_Y \mathcal{L}_X \omega)(Y_1, \dots, Y_{k-1}) = (\mathcal{L}_X \omega) (Y,Y_1, \dots, Y_{k-1}).$$
The global formula for the Lie derivative is 
However, I can't figure out how to get this identity. I would greatly appreciate some help.
I have solved this problem. For convenience, denote $Y_0 := Y$. Applying the global formula to $i_YL_X\omega$ we get $X(\omega(Y_0,Y_1,\dots,Y_{k-1}))-\sum_{i=0}^{k-1}\omega(Y_0, \dots, [X,Y_i],\dots, Y_{k-1}).$
Meanwhile, $(L_X i_Y \omega)(Y_1, \dots, Y_{k-1}) = (L_X (i_Y \omega))(Y_1, \dots , Y_{k-1}) = X(i_Y \omega(Y_1, \dots, Y_{k-1})) - \sum_{i=1}^{k-1} i_Y \omega (Y_1, \dots, [X,Y_i], \dots, Y_{k-1}). $ Now we see that the first terms equate and the second terms vary only by a single term $-\omega([X,Y],\dots, Y_{k-1}),$ which is $-i_{[X,Y]}\omega(Y_1, \dots, Y_{k-1})$.
$L_X\omega(Y_1,...,Y_k)=X.\omega(Y_1,...,Y_k)- \omega([X,Y_1],..,Y_k)-..-\omega(Y_1,..,[X,Y_k])$ implies that
$i_YL_X\omega(Y_1,..,Y_k)=$
$=X.\omega(Y,..,Y_k)-\omega([X,Y],Y_2,..,Y_k)-\omega(Y,[X,Y_2],..,Y_k)-..-\omega(Y,Y_1,..,[X,Y_k])$. ($Y_1$ is replaced by $Y$).
$i_Y\omega(Y_1,..,Y_{k-1})=\omega(Y,Y_1,..,Y_{k-1})$ implies that
$L_Xi_Y\omega(Y_1,..,Y_{k-1})=$
$=X.\omega(Y,Y_1,...,Y_{k-1})-\omega(Y, [X,Y_1],Y_2,..,Y_{k-1})-..-\omega(Y,Y_1,..,[X,Y_{k-1})$.
We deduce that
$L_Xi_Y-i_YL_X=i_{[X,Y]}$.