Multiplication Persistence

[PkmnQ]

(Only calculated up to 4 digits)
Base 2: 11, 1 (1 step)
Base 3: 222, 22, 11, 1 (10 steps)
Base 4: 333, 123, 12, 2 (3 steps)

[GUYTU6J]

Definition?

[Moosey]

11-> 1*1 -> 1 (any base)
222-> 2*2*2-> 22 (base 3) -> 2*2 -> 11 (base 3) ->1.

So in base 10:
999 -> 9*9*9 -> 729 -> 7*2*9 -> 126 -> 12 -> 2



Exponentiation and tetration and pentation, etc. Persistance exist too but are generally either very small or possibly infinitely large.
Example: 222-> 16 -> 1
333-> 19683 -> 1
Likely most numbers have an exponentiation persistence of two to four depending on how long it takes to get a 1 in front.

[toroidalet]

Every number has a finite multiplication persistence, as the product of its digits can be no more than the original number.
For exponentiation persistence, if exponents are evaluated from left to right (1234->((1^2)^3)^4) it ends when either the first digit is 1 or its expansion contains a 0 (or if it ends up with a single digit). If evaluated right to left (1234->1^(2^(3^4))), it ends if the first digit is 1, the second digit that is not a 1 is 0, or one step after the third digit is 0 (as well as some edge cases like 221x which can be enumerated by hand).
Assuming digits of large composite powers (divisible only by 2,3,5 or 7) are randomly distributed, left-to-right exponents should on average halt after reaching 10 digits and right-to-left exponents should halt after less than 3.32 iterations (less because this estimate is based off of Benford's law and doesn't consider the 0 rules).
EDIT: factoring in the second digit 0 rule for right-to-left exponents, we get an average survival time of 2.308 iterations.

[Moosey]

Every number has a finite multiplication persistence, as the product of its digits can be no more than the original number.
For exponentiation persistence, if exponents are evaluated from left to right (1234->((1^2)^3)^4) it ends when either the first digit is 1 or its expansion contains a 0 (or if it ends up with a single digit). If evaluated right to left (1234->1^(2^(3^4))), it ends if the first digit is 1, the second digit that is not a 1 is 0, or one step after the third digit is 0 (as well as some edge cases like 221x which can be enumerated by hand).
Assuming digits of large composite powers (divisible only by 2,3,5 or 7) are randomly distributed, left-to-right exponents should on average halt after reaching 10 digits and right-to-left exponents should halt after less than 3.32 iterations (less because this estimate is based off of Benford's law and doesn't consider the 0 rules).
EDIT: factoring in the second digit 0 rule for right-to-left exponents, we get an average survival time of 2.308 iterations.

What about tetration persistence?

[toroidalet]

All those estimates should be increased by 1, since the last iteration is the one after the first digit is a 1 or the second non-1 digit is a 0 or whatever.

What about tetration persistence?

Right to left and left to right tetration both seem to behave similarly to exponentiation in terms of when they stop, i.e. left-to-right stops after the number contains even a single 0, and so the same figures should apply. (right-to-left should last slightly longer since there are fewer edge cases of the form 221x that lead to it being a single digit).