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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 satisfyTo prove regularity, suppose $\mathrm{cf}({\kappa }^{+})=\mu <{\kappa }^{+}$. Since ${\kappa }^{+}$ is a successor cardinal, $\mu \le \kappa$. Choose a cofinal sequence $⟨{\alpha }_i:i<\mu ⟩$ below ${\kappa }^{+}$. Each ${\alpha }_i$ has cardinal at most $\kappa$, and choice givesBut cofinality makes this union all of ${\kappa }^{+}$, a contradiction. Therefore $\mathrm{cf}({\kappa }^{+})={\kappa }^{+}$, 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.