It is easy to show that for any (Dedekind) infinite cardinal $\kappa$ we have $\aleph_0+\kappa=\kappa$.
Definition of an infinite cardinal is a cardinal such that $\aleph_0\le\kappa$. (I believe this is usually called Dedekind infinite.) We can use basically the same "Hilbert hotel argument" as in the proof of $\aleph_0+\aleph_0=\aleph_0$. This proof does not use Axiom of Choice.
With the use of AC we can show $\aleph_0\cdot\kappa=\kappa$.
We have $\kappa\le\aleph_0\cdot\kappa \le \kappa\cdot\kappa = \kappa$. Then we can use Cantor-Bernstein theorem. In the equality $\kappa\cdot\kappa = \kappa$ we are using Axiom of Choice. (See here, here or here.)
So my question is:
Can $\kappa\cdot\aleph_0=\kappa$ for every $\kappa\ge\aleph_0$ be shown in ZF (i.e., without using Axiom of Choice)?
Disclaimer: I strongly suspect that the answer to this question can be very probably found in some of the other answers on this site or can be derived as a consequence of some result related to choice mentioned in other answers. But I did not find a questions asking explicitly this. I thought it might be useful to have such question as a reference. (If not for other reason, this might help people searching for the answer to this question.)