09/18/2022, 03:13 PM
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.
\[z \uparrow^n \infty = a \implies z = a \uparrow^{n-1} (a \uparrow^{n-1} {-1})\]