= Solution
An <aleph number> is an infinite <initial ordinal>. More explicitly,
$$
\aleph_0=\omega,
\qquad
\aleph_{\alpha+1}=\text{the least cardinal greater than }\aleph_\alpha,
\qquad
\aleph_\lambda=\sup_{\alpha<\lambda}\aleph_\alpha
$$
for a limit ordinal $\lambda$. By the <well-ordering theorem>, every set is equipotent to an ordinal and hence to a unique initial ordinal. If the set is infinite, that initial ordinal occurs in the aleph enumeration. Thus \b[every infinite set has cardinality $\aleph_\alpha$ for a unique ordinal $\alpha$].
We next prove the <square of an infinite cardinal>. Suppose otherwise and let $\kappa$ be the least infinite cardinal for which $\kappa^2>\kappa$. Well-order $\kappa\times\kappa$ by increasing
$$
\max\{\alpha,\beta\},
$$
breaking ties lexicographically. The predecessors of $(\alpha,\beta)$ lie inside $\gamma\times\gamma$ for some $\gamma<\kappa$. If $\mu=|\gamma|$, then $\mu<\kappa$; by minimality, $\mu^2=\mu$ when $\mu$ is infinite, while the finite case is immediate. Every proper initial segment therefore has cardinality less than $\kappa$.
Recursively assign to each pair the least ordinal below $\kappa$ not already assigned to one of its predecessors. Such an ordinal always exists by the preceding bound, so this constructs an injection $\kappa\times\kappa\to\kappa$. The map $\alpha\mapsto(\alpha,0)$ gives the reverse injection, and the <Cantor-Schröder-Bernstein theorem> yields
$$
\boxed{\kappa^2=\kappa.}
$$
For infinite $X_1,\ldots,X_n$, finitely many applications of the <cardinal comparability principle> let us choose $X_i$ of largest cardinality $\kappa$. Then
$$
\kappa
\leq\left|X_1\cup\cdots\cup X_n\right|
\leq |X_1|+\cdots+|X_n|
\leq n\kappa
=\kappa,
$$
where the final equality follows from infinite cardinal arithmetic. Consequently
$$
\boxed{|X_1\cup\cdots\cup X_n|=|X_i|\text{ for some }i.}
$$
For a countable family of pairwise different infinite cardinalities, the answer is \b[yes]. Regard initial ordinals as sets and take
$$
X_1=\aleph_\omega,
\qquad
X_{n+2}=\aleph_n\quad(n<\omega).
$$
The cardinalities are pairwise distinct, but every $X_{n+2}$ is a subset of $X_1$. Hence
$$
\boxed{\bigcup_{j\geq1}X_j=X_1,}
$$
which is the <countable family of distinct infinite cardinalities with a largest member> construction.
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