 Solution

Assume $M\times N$ is oriented. Fix $q\in N$ and an ordered basis $(w_1,\dots ,w_n)$ of $T_{q}N$. Contracting the product orientation successively with the constant vectors $w_i$ along $M\times \{q\}$ produces a nowhere-zero top form on $M$. Hence $M$ is orientable. Fixing a point and a basis in $M$ gives an orientation of $N$ in the same way. This is the orientability of factors of a product manifold.