Stirling Approximation:-
$\large{n! ≈ \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n}$
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$=\large{\lim_{n \to \infty}\dfrac{10^n}{\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n}}$
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$\large{= \lim_{n \to \infty}\left(\dfrac{1}{\sqrt{2\pi n}}\left(\dfrac{10}{\frac{n}{e}}\right)^n\right)}$
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$\large{= \lim_{n \to \infty}\left(\dfrac{1}{\sqrt{2\pi n}}\left(\dfrac{10*e}{n}\right)^n\right)}$
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$\large{n \to \infty,}$ $\large{\dfrac{1}{\sqrt{2\pi n}} \to 0}$, $\large{\dfrac{10*e}{n} \to 0}$
as both limits exists we can you product rule of limits:-
$\large{= \lim_{n \to \infty}\left(\dfrac{1}{\sqrt{2\pi n}}\left(\dfrac{10*e}{n}\right)^n\right)} = \large{= \lim_{n \to \infty}\left(\dfrac{1}{\sqrt{2\pi n}}\right)\lim_{n \to \infty}\left(\dfrac{10*e}{n}\right)^n} = (0).(0)^n = 0$