How many consecutive zeros are there at the end of the integer obtained in the product 1² × 2⁴ × 3⁶ × 4⁸ × … × 25⁵⁰?

The number of trailing zeros equals the power of 5 in the prime factorization (since powers of 2 are abundant). The general term is n^(2n) for n = 1 to 25. We need to count the total power of 5 contributed by each n^(2n). Multiples of 5 in [1,25] are 5, 10, 15, 20, 25. For n=5: 5^(10) contributes 10 fives. For n=10: 10^(20) = 2^20 · 5^20 contributes 20. For n=15: 15^(30) contributes 30. For n=20: 20^(40) = 2^80 · 5^40 contributes 40. For n=25: 25^(50) = 5^(100) contributes 100. Total power of 5 = 10 + 20 + 30 + 40 + 100 = 200. The power of 2 vastly exceeds 200, so the number of trailing zeros equals 200.