= Solution
Define a natural number to be a <Finite von Neumann ordinal>: an ordinal $n$ such that every nonempty subset of $n$ has a greatest member. This avoids the usual impredicative description of $\mathbb N$ as the intersection of all inductive sets.
Every usual von Neumann natural number
$$
0=\varnothing,
\qquad
n+1=n\cup\{n\}
$$
has this property. The proof is by induction: a nonempty subset of $n+1$ either contains $n$, which is then greatest, or is a nonempty subset of $n$.
Conversely, let $\alpha$ be an ordinal with the stated property. If $\alpha$ were not one of the finite von Neumann ordinals, it would contain every finite ordinal. Indeed, if $n$ were the least finite ordinal not in $\alpha$, ordinal comparability and the presence of all $m<n$ would force $\alpha=n$ or $\alpha=m$ for some $m<n$. The subset
$$
\omega=\{0,1,2,\ldots\}\subseteq\alpha
$$
would then be nonempty and have no greatest element, a contradiction. Thus this definition picks out exactly the natural numbers given by the usual least-inductive-set definition.
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