Question

what is the remainder when 7^7^7^upto infinite times(7 to the power 7 to the power 7 to the power ....infinite times) is divided by 13

Answer 1

If we have 7^7^7^7^... then it is obvious an infinity.
So we must have
7^7^7^...^7 only some n-times.
Euler function of 13 is 12. Then 7^12=1(mod
13). 7^7=7(mod 12).
So, starting from tail of 7^7^7^...^7 we can replace
7^7^7 by 7 . So, finally we get the remainder 7 or 7^7=6(mod 13).