If it's about factorials, you can use Stirling's approximation:

Due to the fact that

Error Bound

Writing the "whole" Stirling series as

$$\ln(n!)\approx n\ln(n)−n+\frac{1}{2}\ln(2\pi n)+\frac{1}{12n} −\frac{1}{360n^3}+\frac{1}{1260n^5}+\ldots $$

it is known that the error in truncating the series is always the opposite sign and at most the same magnitude as the first omitted term. Due to Robbins, we can bound:

$$\sqrt{2\pi }n^{n+1/2}e^{-n} e^{\frac{1}{12n+1}} < n! < \sqrt{2\pi }n^{n+1/2} e^{−n} e^{1/12n}$$

More on Stirling Series in Base

Let's develop the question of Stirling series when we have a base for example. The above approximation has to be read this way:

$$log_2(N!) \approx \log_2(\sqrt{2\pi N} N^N\ e^{-N})$$

Due to the fact that we have a non-natural log, it becomes

Hence one has to be very careful with the last term which is not anymore, but .

That being said one can proceed with the rest of Stirling series.

See the comments for numerical results.

Beauty Report

$$\color{red}{256\log_2(256) - 256\log_2(e) + \frac{1}{2}\log_2(2\pi\cdot 256) = 1683.9958175971615}$$

a very good accord with numerical evaluation (for example W. Mathematica) which gives $\log_2(256!) = 1683.9962872242145$.