Integration is all about taking areas in small chunks and summing them together.
$$ \begin{align*} \int^a_bf(x)dx&=f(a)dx+f(a+dx)dx+f(a+2dx)dx+\dots+f(b)dx\\ &=\left(f(a)+f(a+dx)+f(a+2dx)+\dots+f(b)\right)dx\\ &=\lim_{dx\to 0}\sum^{b}_{n=a\atop{\text{step}=dx}}f(a+ndx)dx \end{align*} $$
Instead of contionuously adding all the areas together, what if we continuously multiply the areas together? $$f(a)\approx\sqrt[\frac{1}{dx}\space\space\space]{(f(a)\times f(a+dx)\times f(a+2dx)\times\dots\times f(a+1))}\qquad\text{there are $\frac{1}{dx}$ terms}$$
To ensure that for each unit change in \(x\) there is only 1 value to be multiplied, root everything by \(\frac{1}{dx}\).
$$ \begin{align*} \text{\Large{P}}_{b}^af(x)^{dx}&=(f(a)\times f(a+dx)\times f(a+2dx)\times\dots\times f(b))^{dx}\\ &=\lim_{dx\to 0}\prod^{b}_{n=a\atop{\text{step}=dx}}f(a+ndx)^{dx} \end{align*} $$
Or it could be written in a way that uses existing integration rules.
$$ \begin{align*} &=(f(a)\times f(a+dx)\times f(a+2dx)\times\dots\times f(b))^{dx}\\ &=e^{\ln(f(a))+\ln(f(a+dx))+\ln(f(a+2dx))+\dots+\ln(f(b))dx}\\ &=e^{\int^a_b\ln(f(x))dx} \end{align*} $$
This is mattaplication by Lavaan.
$$ \begin{align*} n!&\approx k\text{\Large{P}}^n_1(x+0.5)^{dx}\\ &=k\text{\Large{P}}^{n+0.5}_{1.5}x^{dx} \end{align*} $$
Correction term \(k\) has been taken from Stirling’s approximation, by trial and error the best value we found is \(k=\sqrt{2\pi}\left(\frac{1.5}{e}\right)^{1.5}\), which simplifies to be
$$n!\approx\sqrt{2\pi}\left(\frac{n+0.5}{e}\right)^{n+0.5}$$
where \(+0.5\) accounts for continuity corrections. The reason why \(k\) takes such a specific value is not known.
This approximation is asymptotic with Stirling’s approximation, when \(n\) is large
$$n+0.5\approx n$$
Therefore
$$ \begin{align*} \sqrt{2\pi}\left(\frac{x+0.5}{e}\right)^{x+0.5}&=\frac{1}{e^{x+0.5}}\sqrt{2\pi(x+0.5)}(x+0.5)^x\\ &\approx\frac{1}{\sqrt{e}}\sqrt{2\pi x}\left(\frac{x}{e}\right)^x\qquad\text{where $\sqrt{2\pi x}\left(\displaystyle\frac{x}{e}\right)^x$ is Stirling’s approximation.} \end{align*} $$
Considering the all the \(+0.5\) removed, their added values can probably cancel out \(\frac{1}{\sqrt{e}}\). Note that this approximation gives values larger than the true value of \(x!\) while Stirling’s approximation (no corrections) gives a smaller value.
Rearranging the equation shows the approximation from mattaplication is similar to a shifted graph of Stirling’s approximation. (except the last exponent \(x\)).
$$ \begin{align*} \text{mattaplication-derived approximation}=&=\sqrt{2\pi(x+0.5)}\cdot\frac{1}{e^{x+0.5}}(x+0.5)^x\\ \text{Stirling’s approximation}&=\sqrt{2\pi x}\cdot\frac{1}{e^x}x^x \end{align*} $$
Expressing Stirling’s approximation with mattaplication
$$ \sqrt{2\pi n}\left(\frac{n}{e}\right)^n=\sqrt{2\pi n}\cdot\frac{1}{e}\cdot\text{\Large{P}}^n_1x^{dx} $$
Why is the correction term not needed anymore? I have no idea.