= Solution
If zero is admitted as a limit case, the assertion there is immediate since $\gamma_0=0$. For a nonzero limit, let $\kappa=|\lambda|$, an infinite <initial ordinal>. There are two cases; separating them avoids choosing an entire family of <bijections> at earlier stages.
If $\kappa<\lambda$, the hypothesis applies at $\kappa$. Part (a) gives a <bijection> $g_\kappa:\kappa^2\to\gamma_\kappa$, whose range has <cardinality> $\kappa$ by hypothesis. Fix one <bijection> $\lambda\to\kappa$; taking its product gives $\lambda^2\cong\kappa^2$. Since $g_\lambda$ is also bijective onto its range, these maps give $\boxed{|\gamma_\lambda|=\kappa=|\lambda|}$.
If $\lambda=\kappa$ is itself an <initial ordinal>, suppose $\gamma_\lambda>\kappa$. Continuity gives $\gamma_\lambda=\bigcup_{\alpha<\kappa}\gamma_\alpha$, so some earlier $\gamma_\alpha$ contains the <ordinal> $\kappa$. If $\alpha$ is finite then $\gamma_\alpha=\alpha^2$ is finite, impossible. If $\alpha$ is infinite, the hypothesis gives $|\gamma_\alpha|=|\alpha|<\kappa$, again impossible because its <ordinal> range contains an initial segment of size $\kappa$.
Thus $\gamma_\lambda\leq\kappa$. Conversely, <transfinite induction> gives $\gamma_\alpha\geq\alpha$: each successor appends a nonempty row and limits take unions. In particular $\gamma_\lambda\geq\lambda=\kappa$. Hence $\gamma_\lambda=\kappa$ in this case, proving the required cardinal equality as well. The argument includes $\lambda=\omega$, whose earlier stages are finite.
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