Which kind product of non-zero number non-zero cardinal numbers yields zero? Let $I$ be a non-empty set. $\kappa_i$ is non-zero cardinal number for all $i \in I$.
If without AC, then $\prod_{i \in I}\kappa_i=0$ seems can be true(despite I still cannot believe it).
But what property should $I$ and $\kappa_i$ have?
Can $\prod_{i \in I}\kappa_i\ne 0$ be proved without AC when $I$ and each $\kappa_i$ all is well-orderable?
Conversely if $I$ is not well-orderable, or if some $\kappa_i$ is not well-orderable, is $\prod_{i \in I}\kappa_i=0$ definitely holds?

The question is based on presuppositions that might not be true in the absence of AC.  Let's consider the simplest non-trivial case, the product of countably many copies of 2, that is, $\prod_{n\in\mathbb N}\kappa_n$ where $\kappa_n=2$ for all $n$.  A reasonable way to define this product would be: Take a sequence of sets $A_n$ of the prescribed cardinalities $\kappa_n$, let $P$ be the set of all functions $f$ that assign to each $n\in\mathbb N$ an element $f(n)\in A_n$, and then define the product to be the cardinality of $P$.  Unfortunately, the cardinality of $P$ can depend on the specific choice of the sets $A_n$.
On the one hand, it is consistent with ZF that there is a sequence of 2-element sets $A_n$ for which there is no choice function; that is, the $P$ defined above is empty.  So these $A_n$'s lead to a value of 0 for the product.
On the other hand, we could take $A_n=\{0,1\}$ for all $n$, and then there are lots of elements in $P$, for example the constant function with value 0.  Indeed, for any subset $X$ of $\mathbb N$, its characteristic function is a member of $P$.  The resulting value for the product of countably many 2's would then be the cardinality of the continuum.
The moral of this story is that, in order for infinite products to be well-defined, one needs AC (or at least some special cases of it), even when the index set and all the factors in the product are well-orderable.
Digging into the problem a bit more deeply, one finds that the natural attempt to prove that "the cardinality of $P$ is independent of the choice of $A_n$'s" involves the following step.  If we have a second choice, say the sets $B_n$, and we know that each $A_n$ has the same cardinality as the corresponding $B_n$, so we know that there are bijections $A_n\to B_n$ for all $n$, then we need to fix such bijections --- to choose a specific such bijection for each $n$.  Then those chosen bijections can be used to define a bijection between the resulting two versions of $P$.  But choosing those bijections is an application of the axiom of  choice. 
