= Solution
The finite-stage calculation gives $\gamma_\omega=\sup_{n<\omega}n^2=\omega$, so the desired equality holds at the first infinite <ordinal>. Apply <transfinite induction> to all subsequent infinite <ordinals>. The successor step is part (b), and the limit step is part (c). Thus
$$
|\gamma_\alpha|=|\alpha|\quad\text{for every infinite ordinal }\alpha.
$$
By the <well-ordering theorem>, any infinite cardinal in the question may be represented by an infinite <initial ordinal> $\kappa$. The corrected recursion supplies a <bijection> $g_\kappa:\kappa\times\kappa\to\gamma_\kappa$, and the just-proved equality gives $|\gamma_\kappa|=\kappa$. Therefore
$$
\boxed{\kappa\cdot\kappa=\kappa.}
$$
The proof used only finite-copy interleaving before reaching this conclusion; using the square equality itself to estimate the successor or limit stages would have been circular.
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