Prime Exponents Tower

Consider the sequence of remainders when $2^n$ is divided by 10. For $n = 1, 2, 3, \ldots$ , we obtain $2, 4, 8, 6, 2, 4, 8, 6, \ldots$ . The pattern repeats every 4 terms. So we want to discover what is the remainder of $3^{5{^\ldots }}$ when divided by 4.

Consider the sequence of remainders when $3^n$ is divided by 4. For $n = 1, 2, 3, \ldots$ , we obtain $3, 1, 3, 1, 3, 1, \ldots$ . The pattern repeats ever 2 terms. So we want to discover what is the remainder of $5^{7^{\ldots } }$ when divided by 2.

The number 5 is odd, so $5^n$ is always going to be odd.

Hence $3^ { 5 ^ \ldots }$ is going to leave a remainder of 3 when divided by 4.

Hence $2 ^ { 3 ^{ 5 ^ { \ldots } } }$ is going to leave a remainder of 8 when divided by 10. Thus, the last digit is 8.

The number $5$ is odd, as is $5^m$ for any natural number $m$ . Hence $3^{5^m}\equiv 3(\text{mod }4)$ , and therefore $2^{3^{5^m}}\equiv 2^3 \equiv 8 (\text{mod }10)$ .

5^(anything) has last digit 5

3^(5) or 3^(25) ... has last digit 3

2^(multiple of 3 having last digit 3) = 8

solution by: Sushil Verma