Why is the cardinality of $\bigcup\limits_{n\in\omega} \underbrace{\aleph_2\times\dots\times\aleph_2}_{n\text{ times}}$ equal to $\aleph_2$?
The cardinality of a set is the smallest ordinal number $\alpha$ such that there is a bijection $\alpha\xrightarrow[]{\text{bij}}\bigcup\limits_{n\in\omega} \underbrace{\aleph_2\times\dots\times\aleph_2}_{n\text{ times}}$
But I have no idea what $\aleph_2$ looks like
$\aleph_2$ is the smallest cardinal that is bigger than $\aleph_1$, which is the smallest cardinal bigger than $\aleph_0=\omega$. We cannot say much beyond that about $\aleph_2$, for example how it compares to the cardinality of $\Bbb R$.
However, if $\alpha$ is any infinite cardinal, then $\alpha\times \alpha\approx \alpha$ by essentially the same diagonal argument that shows $\Bbb N\times \Bbb N\approx \Bbb N$. By induction, $\alpha\times\ldots\times \alpha\approx \alpha$ and even more so $\alpha\times \omega\approx \alpha$ and of course the union of $\omega$-many sets of cardinality $\alpha$ is also $\approx\alpha$.