How good is Kamenetsky's formula for the number of digits in n-factorial? In Number of digits in n!, now closed, there was a mention of Dmitry Kamenetsky's formula, $[\bigl(\log(2\pi n)/2+n(\log n-\log e)\bigr)/\log 10]+1$, for the number of decimal digits in $n$-factorial. Here, $[x]$ is the integer part of $x$. The formula appears at A034886 in the Online Encyclopedia of Integer Sequences, http://oeis.org/A034886. My question is whether this formula is exact for all $n$, or is it occasionally off. No proof of exactness is given at the OEIS, no paper of Kamenetsky appears in Math Reviews. 
In the other thread, I mentioned the discussion in the Usenet newsgroup sci.math in January-February, Subject: Number of digits in factorial. Although neither proof nor counterexample was found, I'd recommend looking over that discussion before starting in on this question. 
EDIT 11 Aug 2011: I note that the question also came up at m.se: question 8323, 30 Oct 2010.
 A: This is a simple computation using the asymptotic formula for $\log_{10}(n!)$.  Computing with $\ln$ instead (just divide the results by $\ln(10)$), Maple gives
$$ \ln{n!} \sim \left(\ln(n)-1\right)n+\ln(\sqrt{2\pi})+\frac{\ln(n)}{2}+\frac{1}{12}n^{-1}-\frac{1}{360}n^{-3}+\frac{1}{260}n^{-5}-\frac{1}{1680}n^{-7}+O(n^{-9})$$
for the expanded version of (the logarithm of) Stirling's formula.  So as long as 1 is larger than the remainder after taking the first 3 terms of the above formula, the formula is quite good.  Only for few $n$ could you run into problems.
I wouldn't be surprised if this question got closed too - it was just too easy to answer using any CAS.

Edit: since I now understand the question better, and Noam Elkies reported a new result in his search, I figured I would try to add one more term to the approximation and see what I get.  More specifically, use 

log10((n/exp(1))^n*sqrt(2*Pi*n)*exp(1/12/n))

(in Maple notation) instead of the original formula.  For $n=6561101970383$, this approximation gives exactly the same digits as displayed in Noam's answer for the exact answer.
In other words, I would conjecture that using this particular approximation, whatever counter-examples there might be would be so large that we may never be able to exhibit them.  Call it an exact approximation for ultra-finitists if you will.
A: Hello all,
My formula is only an approximation, since it uses Stirling's approximation for $n!$. I expect the formula to fail sometimes. I have increased the precision of my program and I can confirm that the formula doesn't fail for $n\leq5\cdot10^7$. I will be running some further tests to see if I can improve this bound. Stay tuned.
