I'm studying differential forms, and in my homework I'm asked to show that the product of two manifolds $M \times N$ is orientable if and only if both $M$ and $N$ are orientable.
I want to show this using volume forms. For the backwards implication ($\Leftarrow$):
Suppose $M$, m-manifold, and $N$, n-manifold, are orientable. Then there exist nowhere vanishing top forms $\omega_1 \in \Omega^m(M)$ and $\omega_2 \in \Omega^n(N)$. Define $\omega_1 \times \omega_2$ in $M \times N$ by $\omega_1 \times \omega_2(X_1,...,X_m,Y_1,...,Y_n)=\omega_1(X_1,...,X_m)\omega_2(Y_1,...,Y_n)$, and one can easily see that this is a form, a top, nowhere vanishing $(m+n)$-form in $M\times N$. It follows that $M\times N$ is orientable.
My problem is in the forward implication. How do I show that if $M\times N$ is orientable, then so are $M$ and $N$, or, equivalently, if $M$ and $N$ are both not orientable, then $M\times N$ can't be orientable? How can I construct volume forms on $M$ and $N$ from a given volume form in $M\times N$?
Assume $M \times N$ is orientable. Fix a point $p \in N$ and a basis $\{v_1, \ldots, v_n\}$ of $T_p N$. Consider an orientation form $\omega$ of $M \times N$ and identify $M$ with $M \times \{p\} \subset M \times N$. Now define $\eta$ on $M$ by $\eta(e_1, \ldots, e_m) = \omega(e_1, \ldots, e_m, v_1, \ldots, v_n)$.
(edited to make argument much simpler)