= Solution
Use <ZFC> and infinite input sets, the usual scope of the stated claim. For a finite set of size $n$, its <Hartogs number> is $n+1$, which is a finite successor cardinal rather than an aleph. Without the <axiom of choice>, one cannot silently identify every infinite input with an aleph, and the asserted conclusion needs those hypotheses.
The <Hartogs number> $h(X)$ is the least ordinal that does not inject into $X$. It is an <initial ordinal>: if it were equinumerous with a smaller ordinal $\alpha<h(X)$, composing that bijection with an injection $\alpha\to X$ would contradict its definition. With choice, well-order an infinite $X$ and put $|X|=\kappa=\aleph_\eta$. Every ordinal below $\kappa^+$ has cardinal at most $\kappa$ and injects into $X$, whereas $\kappa^+$ does not. Hence the <Hartogs numbers under choice> satisfy
$$
\boxed{h(X)=\kappa^+=\aleph_{\eta+1}}.
$$
To prove regularity, suppose $\operatorname{cf}(\kappa^+)=\mu<\kappa^+$. Since $\kappa^+$ is a successor cardinal, $\mu\le\kappa$. Choose a cofinal sequence $\langle\alpha_i:i<\mu\rangle$ below $\kappa^+$. Each $\alpha_i$ has cardinal at most $\kappa$, and choice gives
$$
\left|\bigcup_{i<\mu}\alpha_i\right|\le\mu\cdot\kappa\le\kappa.
$$
But cofinality makes this union all of $\kappa^+$, a contradiction. Therefore $\operatorname{cf}(\kappa^+)=\kappa^+$, and every infinite-input Hartogs value is a regular initial successor aleph. Conversely every successor aleph occurs by taking $X$ to be its predecessor cardinal.
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