I am tackling this problem: "Show that $\Gamma(\alpha\times\alpha)\leq\omega^{\alpha}$". However I'm going on writing pages and pages trying to come up with a reasonable proof, without succeeding.
Here $\Gamma(\alpha,\beta):=$ the order type of the set $\{(\xi,\eta): (\xi,\eta)<(\alpha,\beta)\}$, and $(\alpha,\beta)<(\gamma,\delta)$ if and only if either $max\{\alpha,\beta\}<max\{\gamma,\delta\}$ or $max\{\alpha,\beta\}=max\{\gamma,\delta\} \wedge \alpha<\gamma$ or $max\{\alpha,\beta\}=max\{\gamma,\delta\} \wedge \alpha=\gamma \wedge \beta<\delta$.
I tried to use induction on $\alpha$, but the case of $\alpha$ successor is a mess, and I don't think there could be a simple way out. Therefore I tried to show the fact directly, using the order defined on couples of ordinals (it is really convenient to draw a cartesian plain with the first and the second element of the couples on the axis, realising that the order follows a sort of square-rule). Neither does it seem to work! Anyone coming up with a good idea? I need some help, please, I am running out of ideas.
HINT: Let $\gamma_\alpha=\Gamma(\alpha\times\alpha)$. Then $\gamma_{\alpha+1}=\gamma_\alpha+\color{crimson}{\alpha}+\color{blue}{\alpha+1}=\gamma_\alpha+\alpha\cdot2+1$. (Here $\color{crimson}{\alpha}$ is for the top edge of the enlarged square, and $\color{blue}{\alpha+1}$ is for the righthand edge.) If $\gamma_\alpha\le\omega^\alpha$, then
$$\gamma_{\alpha+1}\le\omega^\alpha+\alpha\cdot2+1\;,$$
and you’d like to show that this is at most $\omega^{\alpha+1}$.
Suppose that you knew that $\alpha\cdot\omega\le\omega^\alpha$; then you’d have
$$\gamma_{\alpha+1}\le\omega^\alpha+\alpha\cdot2+1\le\omega^\alpha+\alpha\cdot\omega\le\omega^\alpha\cdot2\le\omega^{\alpha+1}\;.$$
Try to prove that $\alpha\cdot\omega\le\omega^\alpha$ by induction on $\alpha$.