How to find the unit digit of large powers
Nobody's computing 7²⁵³ by hand — but exams don't need the full number, just its last digit. That's a completely different, much faster problem.
The key idea: last digits repeat in cycles
The last digit of powers of any number repeats in a short, predictable cycle. Once you know the cycle, you only need to know where in the cycle a given exponent lands.
The cycles for each last digit
0, 1, 5, 6 → always end in themselves (cycle length 1)
4, 9 → cycle length 2
2, 3, 7, 8 → cycle length 4
Example — powers of 7: 7¹=...7, 7²=...9, 7³=...3, 7⁴=...1, then 7⁵=...7 again. The pattern (7, 9, 3, 1) repeats every 4 powers.
The method
Find the last digit of the base. Find its cycle length (from the table above). Divide the exponent by the cycle length and look at the remainder — that remainder tells you which position in the cycle to use (with the cycle length itself standing in for a remainder of 0).
Worked example: last digit of 7²⁵³
Last digit of base: 7. Cycle length: 4. 253 ÷ 4 = 63 remainder 1. Remainder 1 means the 1st position in the cycle (7, 9, 3, 1) → 7.
Worked example: last digit of 8¹⁰⁰
Last digit of base: 8. Cycle length: 4 — the pattern is (8, 4, 2, 6). 100 ÷ 4 = 25 remainder 0. A remainder of 0 means the last position in the cycle, not skip it — 6.
The remainder-zero trap
This is where most mistakes happen: a remainder of 0 doesn't mean "nothing" — it means the exponent is an exact multiple of the cycle length, which lines up with the last entry in the cycle, not the first.