Answer: C (0)
The powers of 8 cycle through unit digits 8, 4, 2, 6 (for n = 1, 2, 3, 4) and repeat. Since $8^n = 2^{3n}$, its prime factorisation contains only 2, never 5, so the unit digit 0 is impossible.
The unit digits of powers of 8 follow a cycle of 4: 8 → 4 → 2 → 6 → 8… So digits 2, 4, and 6 all appear. For a number to end in 0, it must be divisible by both 2 and 5; since $8^n = 2^{3n}$ has no factor of 5 (by FTA, prime factorisation is unique), it can never end in 0.