Let $X$ and $Y$ to be two smooth projective algebraic varieties of dimension $n$ and $m$, respectively. Then, $K_{X\times Y}=\operatorname{pr}_1^*K_X+\operatorname{pr}_2^*K_Y$.
Can we can compute $K_{X\times Y}^{n+m}$ in terms of $K_X^n$ and $K_Y^m$?
I'm specially interested in the case of $\dim X = 2$ and $Y\cong \mathbb{P}^1$.
Thanks in advance for any help.
Since $\mathrm{pr}^*$ respects intersection products, we have $\mathrm{pr}_1^*K_X^{n+1} = 0$, since $K_X^{n+1} = 0$ on $X$. Similarly $\mathrm{pr}_2^*K_Y^{m+1} = 0$.
So, expand $K_{X \times Y}^{n+m} = (\mathrm{pr}_1^*K_X + \mathrm{pr}_2^*K_Y)^{n+m}$ as a sum of products. By the above, all the terms vanish except for ${n+m \choose n}\cdot \mathrm{pr}_1^*K_X^n \cdot \mathrm{pr}_2^*K_Y^m$.
Edit: Forgot a binomial coefficient. Also, in the case of $\mathbb{P}^1 \times \mathbb{P}^1$, we have $K = c_1(\mathcal{O}(-2,-2))$, and the intersection product is $8$ times the class of a point.