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For infinite sets in ZFC, the Hartogs number is the successor of the input cardinal. Every smaller ordinal injects into the input, whereas that successor does not. An infinite successor cardinal ${\kappa }^{+}$ is regular: a cofinal family of at most $\kappa$ ordinals below it would have union of cardinal at most $\kappa \cdot \kappa =\kappa$, a contradiction. Finite inputs give finite successor cardinals, so the formulation in terms of alephs requires an infinite input. The general Hartogs theorem itself does not require choice.

For every set $X$ there is a least ordinal $h(X)$ that does not inject into $X$.

The least ordinal supplied by the theorem is the Hartogs number.