Proving $\text{card}(A\times A)=\text{card}(A)$ by transfinite induction?

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The fact that for infinite sets $A,B$, $\text{card}(A\times B)=\max\{\text{card}(A),\text{card}(B)\}$ depends on the axiom of choice or Zorn's lemma, so presumably, it can be proven by transfinite induction. Here is my attempt to prove this by induction.

Note that this is obviously true for $|A|=|B|=\aleph_0$. Now suppose for some cardinal $\kappa$, $\text{card}(A\times B)=\max\{\text{card}(A),\text{card}(B)\}$ is true for all $|A|,|B|<\kappa$. Let $\alpha<\kappa$ be an ordinal. Then by induction assumption, we have $|\alpha\times\alpha|=|\alpha|$. So $$ |\alpha\times \kappa|=|\alpha\times \alpha\times \kappa|=\left|\bigcup_{\beta<\alpha} (\{\beta\}\times \alpha\times \kappa)\right|\\ \left|\alpha\times \kappa\times \bigcup_{\beta<\alpha} (\{\beta\})\right|=|\alpha\times\kappa||\alpha|. $$ But this is very far from the desired $\kappa.\kappa=\kappa$. How can I finish the proof?

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You can write $$\kappa \times \kappa = \left( \bigcup_{\alpha < \kappa} \alpha \right) \times \left( \bigcup_{\beta < \kappa} \beta \right) = \bigcup_{\alpha,\beta < \kappa} \alpha \times \beta \le \bigcup_{\alpha,\beta < \kappa} (\alpha \cup \beta) \times (\alpha \cup \beta)$$ Now apply the induction hypothesis.