1. 7 ≡ 2 (mod 5), so 7¹⁰⁰ ≡ 2¹⁰⁰ (mod 5).
2. Remainders of 2ⁿ on division by 5: 2¹ → 2, 2² → 4, 2³ → 3, 2⁴ → 1, then it repeats every 4.
3. 100 = 4 × 25, so 2¹⁰⁰ sits at the end of a cycle, where the remainder is 1.

Answer: 1.

Reduce the base modulo the divisor first, then look for the cycle in the remainders of successive powers. The exponent's position in that cycle (exponent mod cycle length) gives the answer without any large arithmetic.

Shortcut. The remainder on 5 can be read off the unit digit: 7¹⁰⁰ ends in 1 because 100 is a multiple of 4, so it leaves 1.

Common trap. 2 — treating 100 mod 4 = 0 as the first position of the cycle 2, 4, 3, 1; a zero remainder means the last position.