Is $\Gamma(\alpha,\alpha)\le \alpha^2$ where $\Gamma$ is the Gödel paring function? 
Is $\Gamma(\alpha,\alpha)\le \alpha^2$ where $\Gamma$ is the Gödel paring function (more precisely, $\Gamma(\alpha,\alpha)=\text{Ord}(\alpha\times \alpha)$ under the canonical well-ordering of $ ON \times  ON$)?

The motivation of this question comes from proving that all multiplicatively indecomposable ordinals $\lambda$ are closed under the Gödel paring function. It is natural to mimic the proof of $\Gamma(\omega_\alpha,\omega_\alpha)=\omega_\alpha$ and try to prove an inequality of this sort (because then $\Gamma(\lambda,\lambda)=\sup_{\alpha<\lambda}\Gamma(\alpha,\alpha) \le \sup_{\alpha<\lambda}\alpha^2\le\lambda$). But the obvious avenue of attack, that is, induction, fails at the successor step (since $\Gamma(\alpha+1,\alpha+1)\le\alpha^2+\alpha\cdot2+1\nleq (\alpha+1)^2$). The proof still works if we use the weaker inequality $\Gamma(\alpha,\alpha)\le \alpha^3$ (which can be proved by induction), so this naturally leads to the question of whether $\Gamma(\alpha,\alpha)\le \alpha^2$ holds in general.
 A: This is in fact not true.  Let me abbreviate $\Gamma(\alpha,\alpha)$ as $f(\alpha)$.  I claim first that for any ordinals $\alpha,\delta$ such that $\delta>0$ and $\alpha\geq \omega^{\delta-1}$ (where we say $\delta-1=\delta$ if $\delta$ is a limit ordinal), we have $$f(\alpha+\omega^\delta)=f(\alpha)+\alpha\cdot\omega^\delta.\qquad(*)$$  Let us prove this by induction on $\delta$.  In the base case $\delta=1$, this is because $f(\alpha+\omega)=f(\alpha)+\sum_{n<\omega}((\alpha+n)\cdot 2+1)$ and the latter sum is easily seen to be equal to $\alpha\cdot\omega$.  When $\delta$ is a limit ordinal, the conclusion is immediate because both sides of $(*)$ are continuous in $\delta$.  For the successor case, suppose we know that $(*)$ holds for a certain value $\delta$ and all $\alpha\geq \omega^{\delta-1}$ and we wish to show it also holds for $\delta+1$ and all $\alpha\geq\omega^\delta$.  Note that we then have $$f(\alpha+\omega^\delta\cdot n)=f(\alpha)+\alpha\cdot\omega^\delta+(\alpha+\omega^\delta)\cdot\omega^\delta+\dots+(\alpha+\omega^\delta\cdot(n-1))\cdot\omega^\delta.$$  Since $\alpha\geq\omega^\delta$, $(\alpha+\omega^\delta\cdot i)\cdot\omega^\delta=\alpha\cdot\omega^\delta$ for any $i<\omega$, so this reduces to just $f(\alpha+\omega^\delta\cdot n)=f(\alpha)+\alpha\cdot\omega^\delta\cdot n$.  Taking the limit as $n\to\omega$ then gives $f(\alpha+\omega^{\delta+1})=f(\alpha)+\alpha\cdot\omega^{\delta+1}$, as desired.
In particular, applying $(*)$ with $\delta=\gamma+1$ and $\alpha=\omega^\gamma$, we get $f(\omega^{\gamma+1})=f(\omega^\gamma)+\omega^{\gamma+\gamma+1}$.  If $\delta=\gamma+1$ is an infinite successor ordinal, this tells us that $f(\omega^\delta)=f(\omega^{\gamma+1})\geq \omega^{\gamma+\gamma+1}=(\omega^\delta)^2$.  Taking $\alpha=\omega^\delta+1$, we then have $$f(\alpha)\geq (\omega^\delta)^2+\omega^\delta\cdot 2+1$$ but $$\alpha^2=(\omega^\delta)^2+\omega^\delta+1$$ so $f(\alpha)>\alpha^2$.
(The smallest $\alpha$ of this form is $\omega^{\omega+1}+1$.  I have not checked the details but I believe $(*)$ can be used to show this is in fact the smallest $\alpha$ such that $f(\alpha)>\alpha^2$.)
