Thanks Brent.
I did it this way too but realized it may not be correct.

For example: 3^2^5 has 3 as unit digit.

Using your method;
2^5 has unit digit 2
So, 3^2 has unit digit 9

Am I missing something?

Alternate interpretation:

What is the units digit of 12^[13^(14¹�)]?

Determine the CYCLE of units digits when 12 is raised to increasing powers:
12¹ --> units digit of 2
12² --> units digit of 4
12³ --> units digit of 8
12� --> units digit of 6
12� --> units digit of 2
12� --> units digit of 4.

The results above indicate that the units digit repeat in a CYCLE OF 4:
2, 4, 8, 6...2, 4, 8, 6...2, 4, 8, 6...
Implication:
Every exponent that is a MULTIPLE OF 4 completes a cycle, yielding a UNITS DIGIT OF 6.
From there, the cycle REPEATS:
2, 4, 8...
Thus, every exponent that is ONE MORE THAN A MULTIPLE OF 4 begins a new cycle, yielding a UNITS DIGIT OF 2.

When 13 is raised to a power, the result = (multiple of 4) + 1.
13¹ = 13 = 12 + 1 = (4*1) + 1.
13² = 169 = 168 + 1 = (4*42) + 1.
13³ = 2197 = 2196 + 1 = (4*549) + 1.

Thus:
12^(13^power) = 12^(4a + 1), where a is a nonzero integer.
Since the exponent is one more than a multiple of 4, the units digit will be 2.

Mitch- I think you missed some powers in between.

12^ 13 we can take it as 2^13
2^n= 2,4,8,6.........

for n = 13 (with groups of 4) unit digit will be 2

next step for 2^14
2^n= 2,4,8,6.........
for n=14 (with groups of 4) unit digit will be 4

next step for 5^15
4^n= 4,6,4,6,4,6
for n=15 (with groups of 2) unit digit will be 4


Answer is 4