Stirling's Formula

Theorem
$$n! \sim \sqrt {2 \pi} n^n n^{1/2} e^{-n} = \sqrt {2 \pi n} \left({\frac {n} {e}}\right)^n$$ where $$\sim$$ denotes asymptotically equal.

Proof
Consider the sequence $$\left \langle {d_n} \right \rangle$$ defined as $$d_n = \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n$$.

What we want to do is show that $$\left \langle {d_n} \right \rangle$$ is decreasing.

So we examine the sign of $$d_n - d_{n+1}$$.

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It is otherwise known as Stirling's approximation.