My textbook says that we need the axiom of choice to prove that for every infinite cardinal $\lambda$, $\lambda \times \lambda$ is equipotent with $\lambda$. However, I don't see where AC is used in the proof, which goes as follows:
We prove by transfinite induction. Assume that for every $\mu<\lambda$, $\mu\times\mu$ is equipotent to $\mu$. Define an order relation $\leq_R$ on $\lambda\times \lambda$ as follows:
$$(\beta,\gamma)\leq_R (\beta_1,\gamma_1)$$
if and only if
$\sup(\beta,\gamma)<\sup(\beta_1,\gamma_1)$, or $\sup(\beta,\gamma)=\sup(\beta_1,\gamma_1)$ and $\beta<\beta_1$, or $\sup(\beta,\gamma)=\sup(\beta_1,\gamma_1)$ and $\beta=\beta_1$ and $\gamma\leq \gamma_1$.This order is then a well-ordering on $\lambda\times\lambda$. Therefore there exists an ordinal $\alpha$ and an isomorphism $f$ from $(\alpha, \in)$ into $(\lambda\times\lambda, \leq_R)$. We will show that $\alpha \leq \lambda$. If we assume the contrary, then $f(\lambda)=(\beta_0, \gamma_0)$ for some $\beta_0, \gamma_0$ in $\lambda$, and the restriction of $f$ on $\lambda$ is a bijection from
$\lambda$ onto the set $Y=\{(\beta,\gamma) \mid (\beta, \gamma)<_R (\beta_0, \gamma_0)\}$. Let $\delta_0=\sup(\beta_0, \gamma_0)$, then $\delta_0 <\lambda$, and the cardinality of $\delta_0$ is strictly less than $\lambda$, and therefore, by the induction hypothesis, ${\tt card}(\delta_0 \times \delta_0)={\tt card} (\delta_0)< {\tt card}\lambda$. As $Y$ is included in $\delta_0\times\delta_0$, ${\tt card} (Y) < {\tt card} (\lambda)$, contradiction.
Best Answer
The issue here is with the term "cardinal". If you interpret that as "initial ordinal", then choice is not needed here, and instead choice is used to show that every set is equipotent with a cardinal.
If, however, you correctly understand "cardinal" in the general sense, then the axiom of choice is needed, and indeed is equivalent to this statement. But of course, once you assume choice, you may assume that cardinal refers to initial ordinals again...