The Stirling's approximation you want to use for this is
\ln(n) \simeq n\ln(n) - n
This is applicable for n \gg 1. So, for your example
\omega = \frac{1000!}{500!500!}
Take the logarithm of both sides and we find
\omega = \ln(1000!) - 2\ln(500!)
Which, using Stirling's approximation...
So, today while doing my homework for statistical mechanics I was reading about the quantum linear oscillator in the textbook, "Classical and Statistical Thermodynamics" by Ashley H. Carter. In it, after discussing the quantized energy it says:
"Note that the energies are equally spaced and...