Without axiom of choice, how big can the product of a family be? With AC, we have the König's theorem which says that for two family of cardinals
$\kappa_i$ and $\lambda_i$, if for all $i$, $\kappa_i < \lambda_i$, then
$\sum_{i\in I} \kappa_i < \prod_{i\in I} \lambda_i$.
Without AC, $\prod_{i\in I} \lambda_i$ could be empty. But can its cardinality be anything between $0$ and $\sum_{i\in I} \kappa_i$?
 A: This requires some additional clarification.
In principle, when we talk about sums and products of cardinals, we want to say that these do not depend on the choice of the sets we put into the sum and products. But without the axiom of choice it is quite possible that the infinite sums and products of cardinals are not well-defined to begin with.
We can overcome this by saying that ultimately König's theorem is about ordinals, so we are talking about the sum and products of these specific ordinals, and then the cardinalities of those sums and products.
As such, the sums are always well-orderable, at least assuming that the index set was well-orderable. The products, on the other hand, need not be well-orderable. For example, if $\Bbb R$ cannot be well-ordered, then the only infinite products of ordinals which can be well-ordered are those where $0$ appears, or where co-finitely many of the ordinals are $1$.
Still, we can easily show that if the $\prod\lambda_i$ can be well-ordered, then it is certainly strictly larger than the sum $\sum\kappa_i$, under the assumption that $\kappa_i<\lambda_i$, of course. The reason is simple: the well-order gives you all the choice you needed for the proof of König's theorem.

Note that if you simply require that $\kappa_i$ are ordinals and that $\kappa_i<\lambda_i$, we can get an odd situation, with $\lambda_i=\kappa_i+1$, $I=\omega$, and $\sum\kappa_i=\aleph_1$, $\prod\lambda_i=2^{\aleph_0}$, but the two cardinals are incomparable in that case!
Of course in this case neither $\kappa_i$ nor $\lambda_i$ is a cardinal, so this is really stretching the definitions thin.
