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I can give an intuitive explanation for $\infty! = \sqrt{2\pi}$. If you are willing to allow the equation $$x \cdot x \cdot x \cdot ... = \sqrt{1/x}$$ (which follows by exponentiating $x + x + x + ... = -x/2$) then this ``follows" immediately from Wallis' product $$\frac{2^2 \cdot 4^2 \cdot 6^2 \cdot ...}{1^2 \cdot 3^2 \cdot 5^2 \cdot ...} = \frac{\pi}{2}$$ since the left-hand side is formally $$(2^4 \cdot 2^4 \cdot 2^4 \cdot ...) \cdot (\infty!)^2 = \frac{1}{4} (\infty!)^2.$$