Show that the $n$-th power of the symplectic form $\omega$ is preserved under Hamiltonian flow

Question

Consider a manifold $M$ of dimension $2n$ equipped with a symplectic form $\omega$. An arbitrary function $H \in C^\infty(M)$ induces a unique vector field $\chi_{_H} \in \mathfrak{X}(M)$ such that

$$ \text{d}H = \iota_{\chi_{_H}}(\omega) \tag{1}$$

It is easy to see that the flow of $\chi_{_H}$ preserves both the symplectic structure $\omega$ as well as the level surfaces of $H$.

$$ \mathcal{L}_{\chi_{_H}}(\omega) = \mathcal{L}_{\chi_{_H}}(H) = 0 \tag{2}$$

However, I am quite not able to see why the following equality should hold?

$$ \mathcal{L}_{\chi_{_H}}(\omega^{\ n}) = 0 \tag{3}$$

I believe that is the Liouville's theorem.

Answer 1

$L_{X}(\alpha\wedge \beta)=L_X(\alpha)\wedge\beta+\alpha\wedge(L_X(\beta)$.

This implies that $L_{X_H}(\omega\wedge\omega)=L_{X_H}(\omega)\wedge\omega+\omega\wedge(L_{X_H}(\omega))=0$, and recursively, you obtain that $L_{X_H}(\omega^n)=0$.