Say that $M$ is a smooth $n$-dimensional manifold. Is it true that if we have an orientation form (that is, a smooth nowhere vanishing $n$-form) that $\Lambda^n M = \Lambda^n T^*M$ is trivial?
My thinking is that given $\Lambda^n M$ has dimension 1, perhaps we could then show that the orientation form is a basis of $\Lambda^n M$ which would give that it is trivial. Is this the right way of going about this problem, or is there perhaps an easier way to do it?