Calculating Tetration

[flanakin]

You know that (b^n) % m = (b%m)^n, where %m is the remainder when dividing by m. So if you want to know the last 8 digits then you just need to compute %100000000.

I'm not sure that will work. What you're saying is...

Code:

(3^^3) % 100000000 = (3 % 100000000)^^3
7625597484987 % 100000000 = 3^^3
97484987 = 7625597484987

Am I wrong or is this what you're saying? Ultimately, the left side is what I'm trying to get to, but if I can't calculate the tetration because the number is too big, then that won't get me anywhere.