1^∞ and 1^^∞

1^∞ and 1^^∞

One of my favourite facts is that 1^∞ is indeterminate, while 1^^∞ is 1, which is weird cause in general, tetration grows way faster than exponentiation, for example

2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64

2^^1=2
2^^2=4
2^^3=16
2^^4=65,536
2^^5≈2*10^19728
2^^6≈10^10^19727.78

and another reason 1^∞ being indeterminate is weird is because 1^anything=1, even 1^(1089+3318i) is 1, so why is 1^∞ indeterminate, and if it is, shouldn't 1^^∞ be indeterminate?

To explain why 1^∞ is indeterminate, we need to know about limits.

The thing about doing functions with infinity is that you can't just do them, you have to use bigger and bigger numbers and see what it approaches.

The limit of 1^x, as x approaches ∞ is indeed 1, but the limit of (1+(1/x))^x as x approaches ∞ is a form of 1^∞, but this time it's e,

and if you were to take any real positive value smaller than 1 and take it to the power of ∞, it would equal 0, and if you were to take a real positive value above 1 and take that to the power of ∞, it would equal ∞, and limits have to stay the same any way you approach it, so 1^∞ is indeterminate because multiple values can be assigned, it isn't undefined like 1/0 because it has a value, but it has too much values.

Ok so if 1^∞ is indeterminate, why isn't 1^^∞ indeterminate?

This is because unlike infinite exponentiation, infinite tetration has a range of values from e^(-e) to e^(1/e) that can output a real value, for example, (√2)^^∞=2, so you can now approach 1 this time!