Given two manifolds, $M_1 \in \mathbb{R}^{n_1}$ and $M_2 \in \mathbb{R}^{n_2}$, with respective volume forms $\mu_1,\mu_2$ prove that $\mu = \phi_1 ^* \mu_1 \wedge \phi_2 ^* \mu_2$ is a volume form on $M = M_1 \times M_2$.
Here $M$ is a manifold, $\phi_i$ is the projection of $M$ onto $M_i$, e.g. $\phi_1(x_1,x_2) = x_1$.
I know that I can write: $\mu(x,h,k) = \begin{vmatrix} \phi_1 ^* \mu_1(x,h) & \phi_1 ^* \mu_1(x,k)\\ \phi_2 ^* \mu_2(x,h)& \phi_2 ^* \mu_2(x,k) \end{vmatrix}$, now I have to prove that $\mu$ is a volume form but don't know how. Any help would be apprecieated!