= Solution
Call a set function $f$ with <ordinal> domain $\gamma$ good if $f(\xi)=G(\xi,\operatorname{ran}(f|_\xi))$ for every $\xi<\gamma$. Any two good functions agree on their common domain: otherwise their first disagreement has identical earlier ranges and hence identical values under $G$. This is <transfinite induction>, not an assumption of the desired recursion.
Good functions exist with every <ordinal> domain. At a successor stage extend the unique previous function by its prescribed value under $G$. At a limit stage, <Axiom schema of replacement> collects the unique good functions on all smaller domains into a set, and their compatible union is a good function on the limit domain. Each earlier range is a set, so $G$ is applied only to sets. <Transfinite induction> now establishes existence at all stages. Define $F(\alpha)$ to be the value at $\alpha$ of the unique good function on $\alpha+1$. This is a definable class function because goodness and the uniqueness assertion are formulas in the defining formula for $G$. Compatibility gives
$$
\boxed{F(\alpha)=G(\alpha,\{F(\beta):\beta<\alpha\}).}
$$
An <initial ordinal> is an <ordinal> not bijective with any smaller <ordinal>. It is the canonical <ordinal> representative of a well-orderable <cardinal>, so <cardinal arithmetic> can be defined by taking the <initial ordinal> equinumerous with a disjoint union, Cartesian product or function set. In the usual setting with the <axiom of choice>, every cardinal has such a representative. <Hartogs theorem> says that for every set $A$ there is an <ordinal> $h(A)$ which does not inject into $A$, and there is a least such <ordinal>. For an <initial ordinal> $\kappa$ it is the next larger <initial ordinal>: every smaller <ordinal> injects into $\kappa$, and if $h(\kappa)$ were equinumerous with a smaller <ordinal> it too would inject into $\kappa$.
Use the recursion just proved to set $\omega_0=\omega$, $\omega_{\beta+1}=h(\omega_\beta)$ and $\omega_\lambda=\sup_{\beta<\lambda}\omega_\beta$ for nonzero limit $\lambda$. These instructions use the earlier value set alone: at a successor it has a maximum, and at a limit its supremum is taken. They are therefore implemented by an appropriate definable $G$. The limit supremum is initial: if it were bijective with $\gamma$ below it, choose an earlier $\omega_\beta>\gamma$; then $\omega_\beta$ would inject into $\gamma$, contradicting initiality. The sequence is strictly increasing and continuous. There are no omitted infinite <initial ordinals>, since each successor chooses the least next one and each limit is the supremum; more formally, a least omitted <initial ordinal> cannot lie between successive terms, and the least term above it cannot be a limit term. Terms eventually exceed any prescribed <ordinal>, since a strictly increasing <ordinal> sequence satisfies $\omega_\beta\geq\beta$.
There is a <fixed point of a normal ordinal function>. Explicitly let $\alpha_0=0$, $\alpha_{n+1}=\omega_{\alpha_n}$ and $\alpha=\sup_{n<\omega}\alpha_n$. The sequence is strictly increasing, by induction from $0<\omega_0$ and strict increase of the enumeration. It is cofinal in the limit $\alpha$, so continuity gives
$$
\boxed{\omega_\alpha=\sup_n\omega_{\alpha_n}=\sup_n\alpha_{n+1}=\alpha.}
$$
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