Let $X$ and $Y$ be Riemannian manifolds, and let $\omega \in \Omega^i(X)$ and $\eta \in \Omega^j(Y)$ be differential forms on $X$ and $Y$. I expect it is the case that $$ *_{X\times Y}(\omega \wedge \eta) = (*_X \omega) \wedge (*_Y \eta)\,, $$ where '$*$' denotes the Hodge star operation, and where in the expression $\omega \wedge \eta \in \Omega^{i+j}(X\times Y)$ we understand $\omega$ and $\eta$ to be pulled-back from $X$ and $Y$ under projection maps.
The proof is likely quite straightforward, but I found it a bit fiddly, and so I am looking for a reference for this fact, or alternatively a nice proof.
Denote the volume form on $X$ by $\mu_X$ and similarly for $Y,X \times Y$. By the defining property of the Hodge star, we have for all $\alpha \in \Omega^i(X), \beta \in \Omega^j(Y)$
$$ (\alpha \wedge \beta) \wedge \star_{X \times Y} \left(\omega \wedge \eta \right) = \left< \alpha \wedge \beta, \omega \wedge \eta \right>_{X \times Y} \mu_{X \times Y} = \left< \alpha, \omega \right>_X \left<\beta, \eta \right>_Y \mu_X \wedge \mu_Y \\ = \left( \alpha \wedge \star_X \omega \right) \wedge (\beta \wedge \star_Y \eta) = (\alpha \wedge \beta) \wedge \left( (-1)^{|\beta||\star_X \omega|} \star_X \omega \wedge \star_Y \eta \right) \\ (\alpha \wedge \beta) \wedge \left( (-1)^{j(\dim X - i)} \star_X \omega \wedge \star_Y \eta \right). $$
In the calculation above, we used the fact that $\mu_{X \times Y} = \mu_X \wedge \mu_Y$ (this is the definition of the orientation of a product in terms of the orientations on the factors) and $\left< \alpha \wedge \beta, \omega \wedge \eta \right>_{X \times Y} = \left< \alpha, \beta \right>_X \left< \omega, \eta \right>_Y$ which is true because the tangent spaces to the fibers $X \times \{y\}$ and the tangent spaces to the fibers $\{ x \} \times Y$ are orthogonal inside $X \times Y$.
Since the identity above is true for all $\alpha,\beta$, by the non-degeneracy of the wedge product we must have
$$ \star_{X \times Y}(\omega \wedge \eta) = (-1)^{j(\dim X - i)} \star_X \omega \wedge \star_Y \eta. $$