Why is the limit of the factorial function divided by stirling's approximation the 12th root of e?

I was interested in the limit definition of $$e$$, where as $$x$$ tends towards infinity, the limit of $$1$$ plus $$x$$ to the negative first power, all to the power of $$x$$, is in fact $$e$$. I was wondering if the limit of the ratio between the factorial function and Stirling's approximation would converge in a similar manner. The formula in question and the Wolfram Alpha result are: $$\lim_{x\to\infty} (\frac {x!}{\sqrt{2\pi x}(\frac{x}{e})^x})^x = \sqrt[12]{e}$$

Wolfram Alpha evaluated this limit as being the 12th root of $$e$$. My question being, why is this? I understand that this would normally be equal to $$e$$ itself, but what is it about the difference between this approximation and the actual factorial function that takes $$e$$ to the 12th root? Why 12 of all roots? Why not 13? or 14? I'm sure there is an explanation for this, I am just unsure why.

It comes from the next order Stirling approximant. To that order, $$x!=e^{x \ln(x) - x + \frac{1}{2} \ln(2\pi x) + \frac{1}{12x}+o(1/x)}$$. The denominator cancels the first three terms in the exponent, so you're left with $$e^{\frac{1}{12x}}$$ to leading order, and raising that to the $$x$$ gives the result.
This is actually related to the $$(1+1/x)^x$$ definition of $$e$$, in that the inside of the parentheses behaves to leading order like $$1+1/(12x)$$.