Im looking for a detailed proof for $\kappa \cdot \kappa = \kappa $ with $\kappa $ beeing a infinite cardinal number. The problem is described in a book (Frank R. Drake, Set Theory: An Introduction to Large Cardinals) as a exercise but without any solution.
As far as i know i can proof that $\kappa \cdot \kappa = \kappa $ holds for the cardinal $\aleph_0$($\mathbb{N}$) wich can be proven with "Cantor's Diagonal Proof". But i need it for any infinite cardinal so it should also be valid for $\aleph_1,\aleph_2,...$
So im missing the induction part.
I know that it should be possible cause very roughly explained to show something like $| \mathbb{N} \cdot \mathbb{R}| = |\mathbb{R}|$. We can make a bijection if we put all the $\mathbb{N}$ numbers in between 0,1 of the $\mathbb{R}$ and still have enough space to adress all $\mathbb{R}$.
TL;DR im looking for the proof of $\kappa \cdot \kappa = \kappa $ and $\kappa \cdot \lambda = max(\kappa, \lambda) $
$\kappa\lambda = \max\{\kappa, \lambda\}$ is a consequence of $\kappa\kappa = \kappa$.
Indeed, say for instance $\kappa \geq \lambda$. Then for $\lambda\neq 0$, $\kappa \leq \kappa\lambda\leq \kappa\kappa = \kappa$, so $\kappa\lambda= \kappa$.
Then for $\kappa\kappa= \kappa$, you may want to look at my answer here for instance. The idea is a transfinite induction where you well-order $\kappa\times\kappa$ nicely and show that its order type must be $\kappa$ (because each proper initial segment is $<\kappa$)