Idempotent law of cardinal arithmetic I'm reading a (surprisingly short) proof of $\pi \times \pi = \pi$ if $\pi$ is an infinite cardinal from nLab (Theorem 3.1). (I call it surprising because I know that a traditional proof uses Zorn's lemma and is very long.)
However, I can't figure out where the proof uses the condition of infinite cardinal. And I can't see why the last line ``regarding $\alpha$ as an ordinal less than $\pi$'' is true (clearly $\alpha \leq \pi$ but why $\alpha \neq \pi?$).
I would be extremely appreciative of any assistance!
 A: When the proof says

Thus $|\alpha|^2=|\alpha|$ for all ordinals $\alpha<\pi$.

it should state this only for infinite ordinals $\alpha<\pi$.  This is because the statement being proved only applies to infinite cardinals, so assuming $\pi$ is a minimal counterexample only tells you that $\kappa^2=\kappa$ for all infinite cardinals $\kappa<\pi$.
So then, in the final step, in order to say that $|\alpha|^2=|\alpha|<|\pi|$, you need to know that $\alpha$ is infinite.  Also, in order to say $|S(\alpha,\alpha,\alpha)|=|\alpha|^2$, you need to know that $\alpha$ is infinite.  Indeed, $S(\alpha,\alpha,\alpha)$ is in bijection with $(\alpha+1)\times(\alpha+1)\setminus\{(\alpha,\alpha)\}$ by mapping $(x,y,z)\in S(\alpha,\alpha,\alpha)$ to $(y,z)$.  When $\alpha$ is infinite, $|\alpha+1|=|\alpha|$ and so $S(\alpha,\alpha,\alpha)$ is in bijection with $\alpha^2$ with one element removed (which again does not change the cardinality since it is infinite), but this does not work if $\alpha$ is finite.
So, if $\alpha$ is finite, an alternate argument is needed. Assuming $\pi$ is infinite you can instead just say that if $\alpha$ is finite then $|\alpha|^2$ is finite and thus less than $|\pi|$.  It is this alternate argument which fails if $\pi$ is finite.
(As for why $\alpha<\pi$, this is just because $(\alpha,\beta,\gamma)$ is some element of $\pi^3$ which means $\alpha\in\pi$, i.e. $\alpha<\pi$.)
A: FYI. Here is a different proof, by transfinite induction. Let $p$ be an infinite cardinal ordinal. Suppose that $|q\times q|<p$ for every  ordinal $q\in p.$
We use the fact that if $<_W$ is a well-order on a set $A$ and if $|B|<p$ for every $<_W$-initial segment $B$ of $A$ then $|A|\le p.$
For $(x,y), (x',y')\in p\times p$ let $(x,y)<_W (x',y')$ iff
$(i).$ $\max (x,y)<\max (x',y')$ OR
$(ii).$ $\max (x,y)=\max (x',y')$ and $x<x'$  OR
$(iii)$. $\max (x,y)=\max (x',y')$ and $x=x'$ and $y<y'.$
That is, in $(ii)$ and $(iii)$, if $\max (x,y)=\max (x,y')$ then $(x,y), (x',y')$ are $<_W$-ordered lexicographically.
Confirm that $<_W$ is a well-order.
Now if $(x',y')\in p\times p$ and if $B=\{z:z<_W (x',y')\}$ is a $<_W$-initial segment, let $q=\max (x',y')+1$. Then $q\in p$ and  $B\subseteq q\times q ,$ so $|B|\le |q\times q|< p.$
Therefore $[\,\forall q\in p\,(|q\times q|<p) \,] \implies |p\times p|=p.$
Now if there existed an infinite cardinal ordinal $p$ such that $|p\times p|>p$ then consider the $least$ such $p.$ The  reasoning above shows there must exist some $q\in p$ with |$q\times q|\ge p,$ so $|q|$ cannot be a finite cardinal. But then  $|(|q|\times |q|)|=|q\times q|\ge p >|q|$ contradicts the minimality of $p.$
