Dimension of product variety using finite morphisms

Question

I've read the following proof for $\dim(X\times Y)=\dim(X)+\dim(Y)$ with $X,Y$ algebraic varieties. Because dimension is a birrational property we can suppose that $X,Y$ are affine. Now, Noether normalization theorem implies that exists finite surjective maps $f:X\twoheadrightarrow \mathbb{A}^n,g:Y\twoheadrightarrow\mathbb{A}^m$. Then we have a surjective map $f\times g:X\times Y\twoheadrightarrow\mathbb{A}^{n+m}$. As srujective finite morphisms preserves dimension, it is sufficient proof $f\times g$ is a finite morphism, but I don´t know why it is finite. Recall that in the affine case, a morphism $f:X\to Y$ is finite iff the ring of regular functions $\mathscr{O}(X)$ is integral over $\mathscr{O}(Y)$ via the pullback $f^*:\mathscr{O}(Y)\to\mathscr{O}(X)$, ie, if $\mathscr{O}(X)$ is a finitely generated $\mathscr{O}(Y)-$module.