Liouville's theorem

Step 1 (Cartan's magic formula). For any vector field $X$ and form $\omega$, Cartan's formula gives $\mathcal{L}_X\omega=d(\iota_X\omega)+\iota_X(d\omega)$. Apply this to $X$ equal to $X_H$: since $\omega$ is closed, $d\omega=0$, and since $\iota_{X_H}\omega=dH$ by definition of the Hamiltonian vector field, we get $\mathcal{L}_{X_H}\omega=d(dH)+0$.

Step 2 (the form itself does not change). The exterior derivative satisfies $d\circ d=0$ for every function, so $d(dH)=0$. Hence $\mathcal{L}_{X_H}\omega=0$: the Hamiltonian flow preserves the symplectic form itself, not merely its volume.

Step 3 (pass to the top power). The volume form is $\omega^n=\omega\wedge\cdots\wedge\omega$ ($n$ factors). The Leibniz rule for the Lie derivative on wedge products gives $\mathcal{L}_{X_H}(\omega^n)=n\,\omega^{n-1}\wedge\mathcal{L}_{X_H}\omega$, and since $\mathcal{L}_{X_H}\omega=0$ by Step 2, the right-hand side vanishes: $\mathcal{L}_{X_H}(\omega^n)=0$.

Step 4 (integrate along the flow). If $\phi_t$ is the flow of $X_H$, then $\mathcal{L}_{X_H}(\omega^n)=0$ means exactly $\frac{d}{dt}\phi_t^{*}(\omega^n)=0$ for every $t$. Since $\phi_0=\mathrm{id}$ gives $\phi_0^{*}(\omega^n)=\omega^n$, integrating shows $\phi_t^{*}(\omega^n)=\omega^n$ for all $t$, which is Liouville's theorem.