(2k)! must have at least three more factors of 5 within it that twice k! (the same goes for factors of 2 as well, but that will be achieved much earlier).
If k=63, 63 has 14 factors of 5 (floor(63/5) + floor(63/25)).
2*63 = 126, 126 has 31 factors of 5 (floor(126/55) + floor(126/25) + floor (126/125)).
For any smaller k (2k)!/(2*k!) is less than 3.