How to Find the Last Digit of a Huge Power Like 7^100

Updated 12 October 2026

Find the last digit of huge powers like 7^100

Last digit of $7^{100}$7^100 = 1

How do you find the last digit of a huge power like 7^100?

The last digit of a power repeats on a short cycle, so divide the exponent by the cycle length and use the remainder. Powers of 7 end in 7, 9, 3, 1, and then the pattern starts again, because only the last digit of each product matters: 7 × 7 = 49, 9 × 7 = 63, 3 × 7 = 21, and 1 × 7 = 7 brings back the start. The cycle is four long. For 7^100, divide 100 by 4: that is exactly 25 cycles with nothing left over, so 7^100 ends where each cycle ends, in 1. For 7^102, the remainder is 2, so it ends in the second digit of the cycle, which is 9. Every last digit settles into a cycle of length 1, 2 or 4. Powers of 2 end in 2, 4, 8, 6, so 2^50 leaves remainder 2 and ends in 4. Powers of 9 end in 9, 1, so every even power of 9 ends in 1.

1. Powers of 7 walk the same ring of four, however large the exponent gets.

2. 100 is exactly 25 laps of 4, with no steps left over at the end.

3. A whole number of laps finishes on the last stop, so $7^{100}$7¹⁰⁰ ends in 1.

All 36 shortcuts: the Magic Tricks library.