Edit: Posting while asleep

Generalizing \[^\infty z=a \implies z = a^{\frac{1}{a} }\]

gives the base associated with the fixed point for the hyperoperators.

\[z \uparrow^n \infty = a \implies z = a \uparrow^{n-1} (a \uparrow^{n-1} {-1})\]

Generalizing \[^\infty z=a \implies z = a^{\frac{1}{a} }\]

gives the base associated with the fixed point for the hyperoperators.

\[z \uparrow^n \infty = a \implies z = a \uparrow^{n-1} (a \uparrow^{n-1} {-1})\]

Daniel