8^(7^(6^(5^(4^(3^(2^1))))))

[(a)^b implies 'a' raised to the power of 'b', ((a)^b)^c implies 'a' raised to the power 'bc', but a^(b^c) implies 'a' raised to the power 'b' raised to the power 'c'.]

8² mod 10 = 4
8³ mod 10 = 2
8^4 mod 10 =6
8^5 mod 10 = 8. So, it is cyclic with an order of 4 (8,4,2,6)
So, we have to find (7^(6^(5^(4^(3^(2^1)))))) mod 4
which nothing but 3^(6^(5^(4^(3^(2^1)))))) mod 4
But according to Fermat's Little Theorem 3³ mod 4 =1
We can say that (6^(5^(4^(3^(2^1)))))) is a multiple of 3 as
6 is a multiple of 3.
So, (7^(6^(5^(4^(3^(2^1)))))) mod 4 = 1.
which implies that the last digit of 
8^(7^(6^(5^(4^(3^(2^1)))))) is 8.