Pattern in number of digits in $(10^n)!$
Your first link gives an approximation
a(n) = 10^n*(n - 1/log(10)) + n/2 + O(1). [Arkadiusz Wesolowski, Jan 21 2012]
Note that $1-\frac{1}{\log_e(10)} \approx 0.5657055180967481723488710810833949177056\ldots$ so this gives the pattern you are observing
Compare this to $a(40) = 395657055180967481723488710810833949177077$