User blog comment:B1mb0w/Fundamental Sequences/@comment-5529393-20151113121456/@comment-1605058-20151114080557

Let $$\gamma=\varphi(\alpha,\beta)$$. Let me show by induction that $$\gamma\uparrow\uparrow n\leq\omega^{...^{\omega^{\gamma+1}}}$$ with $$n$$ times $$\omega$$.

For $$n=1$$ we clearly have \gamma\uparrow\uparrow 1=\gamma\leq\omega^{\gamma+1}. This is the base of induction. For the step, we have $$\gamma\uparrow\uparrow n+1=\gamma^{\gamma\uparrow\uparrow n}\leq(\omega^\gamma)^{\gamma\uparrow\uparrow n}=\omega^{\gamma\cdot\gamma\uparrow\uparrow n}=\omega^{\gamma\uparrow\uparrow n}\leq\omega^{\omega^{...^{\omega^{\gamma+1}}}}$$ now with $$n+1$$ times $$\omega$$.

Given that, it is now easy to see that if $$\gamma=\varepsilon_\delta$$, then $$\gamma\uparrow\uparrow\omega=\varepsilon_{\delta+1}$$.