At the top of p. 158 of your book, the authors write the following:
Infinite products are more difficult to evaluate than infinite sums. In some special cases ... some simple rules can be proved.
What they're telling you is that you shouldn't expect a simple formula that applies to all infinite products like the one they give for infinite sums. And this is not so surprising: repeated cardinal addition is related to cardinal multiplication, which is a very simple operation. But repeated cardinal multiplication is related to cardinal exponentiation, which is extremely complicated.
In the comments, you made the reasonable guess that $$\prod_{\alpha<\lambda} \kappa_\alpha = (\sup\{\kappa_\alpha\mid \alpha<\lambda\})^\lambda.$$
The formula for infinite sums assumes that all of the $\kappa_\alpha$ are non-zero. To have a hope of the above formula holding, we should assume that all of the $\kappa_\alpha$ are not equal to $0$ or $1$. This handles silly counterexamples like the one in Max's comment. And it's not so bad, because we can remove all $1$ terms from the product without changing its value, while if a single $0$ term appears, the whole product is $0$.
Ok, your guess is reasonable because we have an obvious upper bound: $$\prod_{\alpha<\lambda} \kappa_\alpha \leq (\sup\{\kappa_\alpha\mid \alpha<\lambda\})^\lambda.$$ And equality is achieved sometimes, e.g. as computed on the same page of your book: $$\prod_{n<\aleph_0} n = 2^{\aleph_0} = \aleph_0^{\aleph_0} = (\sup\{n\mid n<\aleph_0\})^{\aleph_0}.$$
On the other hand, it can fail, e.g. assume the continuum hypothesis (we actually only need $2^{\aleph_0} < \aleph_\omega$), let $\kappa_0 = \aleph_\omega$, and let $\kappa_n = 2$ for all $1\leq n<\aleph_0$:
$$\prod_{n<\aleph_0} \kappa_n = \aleph_\omega \cdot \prod_{1\leq n<\aleph_0} 2 = \aleph_\omega \cdot 2^{\aleph_0} = \aleph_\omega < (\aleph_\omega)^{\aleph_0} = (\sup\{\kappa_n\mid n<\aleph_0\})^{\aleph_0}.$$
The fact that $\aleph_\omega < (\aleph_\omega)^{\aleph_0}$ follows from König's Theorem: $$\aleph_\omega = \sum_{n< \aleph_0} \aleph_n < \prod_{n<\aleph_0} \aleph_\omega = (\aleph_\omega)^{\aleph_0}.$$
If you don't want to assume the continuum hypothesis, you can replace $\aleph_\omega$ with any cardinal $\kappa>2^{\aleph_0}$ such that $\text{cf}(\kappa) = \aleph_0$ (if $2^{\aleph_0} = \aleph_\alpha$, then $\kappa = \aleph_{\alpha+\omega}$ works), and the same argument goes through.