How many bits are in factorial? I am interested in good integer approximation from below and from above for binary Log(N!). The question and the question provides only a general idea but not exact values.
In other words I need integers A and B so that  A <= Log(N!) <= B

Expanding on joriki's answer, taking more terms from the approximation,
$$ \log (n!) = n(\log n - \log e) + \frac{1}{2}\log n + \log \sqrt{2\pi} + \frac{\log e}{C_ n}, \quad 12n < C_n < 12n+1. $$
The number of binary digits is equal to $\lceil \log n! \rceil$, and for most $n$, I expect that the slight uncertainty in $C_n$ won't effect $\lceil \log n! \rceil$.
