From n! up to n^n there are only n terms, but this is not the most remarkable

Consider the formula:

$n!=n^n-C_n^1(n-1)^n+C_n^2(n-2)^n-C_n^3(n-3)^n+...+(-1)^{( n-1)}C{_n}^{(n-1)}$

n! = \sqrt{2\pi n}(\frac{n}{e})^n(1+\frac{1}{12 n}+\frac{1}{288n^2}-\frac{139} {51840 n^3}+…)

-\delta(n)=n!-n^n

$We get the formula:[\frac{\sqrt{2\pi n}(1+\frac{1}{12 n}+\frac{1}{288n^2}-\frac{139}{51840 n^3}+...)}{e^n-\sqrt{2\pi n}(1+\frac{1}{12 n}+\frac{1}{288n^2}-\frac{139}{51840 n^3}+...)}] n! \approx \delta(n)$

(**)

$20!-20^{20}=-104857597567097991823360000$

twenty! \approx 2.4329028230918*10^{18}

In this case, the exact value is: 20!=2432902008176640000

A formula of the form (**) shows the relationship between the Stirling series and the series of n terms from the right side of the formula

n!=(\frac{n}{e})^n \sqrt{2 \pi n}\sum_{k=0}^{\infty}\frac{a_k}{n^k} $a_k=\sum_{j=1}^{2k}(-1)^j\frac{d_3(2k+2j,j)}{2^{j+k}(j+k)!},k \in N, (***)$ where