Let ($M$, $\omega$) be a symplectic manifold of dimension $2n$. Then $\omega$ is non-degenerate $2-form$ by definition.
Now, my question is if we can conclude that $\omega \wedge ...\wedge \omega$ is a volume form? (Here the wedge product of $\omega$ is taken $n$ times.)
From my feeling I would say yes, as the dimension fit and maybe the non-degenerateness should give us that this $2n$-form vanishes nowhere, but I don't really know how to show it.
Let $p$ be a point on the manifold. Then there is a coordinate chart $x^1,\ldots,x^n,y^1,\ldots,y^n$, around $p$, such that at $p$ we have$$\omega=\sum dx^i\wedge dy^i.$$(In fact, this equality can always hold on a neighborhood, but we don't need it now). As verified by a straightforward calculation with these coordinates, $\omega^n$ does not vanish at $p$. Since $p$ is arbitrary, $\omega^n$ is a volume form.