User blog comment:B1mb0w/Fundamental Sequences/@comment-5529393-20151113121456/@comment-1605058-20151115092527

Actually we only have $$\varphi(\alpha,\beta)\uparrow\uparrow\omega=\varepsilon_{\varphi(\alpha,\beta)+1}$$ only if $$\alpha\geq 2$$, because then we have $$\varphi(\alpha,\beta)\uparrow\uparrow\omega=\varepsilon_{\varphi(\alpha,\beta)}\uparrow\uparrow\omega=\varepsilon_{\varphi(\alpha,\beta)+1}$$, since for $$\alpha\geq 2$$ we have that $$\gamma=\varphi(\alpha,\beta)$$ satisfies $$\gamma=\varepsilon_\gamma$$.

For $$\alpha=1$$, we indeed have $$\varphi(\alpha,\beta)\uparrow\uparrow\omega=\varphi(\alpha,\beta+1)$$, but this is the only case when it's true.

As for $$\varepsilon_{\zeta_0}$$, it is well defined. Indeed, $$\zeta_0$$ is defined as the least ordinal $$\gamma$$ such that $$\gamma=\varepsilon_\gamma$$. Hence $$\varepsilon_{\zeta_0}=\zeta_0$$.

Indeed, $$\varepsilon_\gamma$$ is defined for every ordinal $$\gamma$$, even for uncountable ordinals.