3 meant to put in a line, not a break

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.

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.

1

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.