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My question is:

$10^8! > 10^{10^9}$ ?

I know that factorial is greater than exponential, but I am not sure about this specific case.

Thanks,

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If we expand $10^8!$, we have the product of $10^8$ positive numbers, all of which are less than or equal to $10^8$. So $10^8!<(10^8)^{10^8}=10^{8\cdot 10^8}<10^{10^9}$.

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$10^8!=\prod \limits_{n=1}^{10^8}n<\prod \limits_{n=1}^{10^8}10^8=(10^8)^{10^8}=10^{8\times10^8}<10^{10\times10^8}=(10)^{10^9}$

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By Stirling’s approximation,

$$\begin{align*} 10^8!&\approx\sqrt{2\cdot10^8\pi}\left(\frac{10^8}e\right)^{10^8}\\ &<10^5\cdot\left(10^8\right)^{10^8}\\ &<\left(10^8\right)^{10^8+1}\\ &=10^{8\cdot10^8+8}\\ &<10^{10^9}\;. \end{align*}$$

Alternatively,

$$\begin{align*} \ln 10^8!&=\sum_{k=1}^{10^8}\ln k\\ &<\int_1^{10^8}\ln x\,dx\\ &=\left[x\ln x-x\right]_1^{10^8}\\ &=8\cdot10^8\ln 10-10^8+1\\ &<10^9\ln 10\\ &=\ln 10^{10^9}\;, \end{align*}$$

and $f(x)=\ln x$ is strictly increasing, so $10^8!<10^{10^9}$.

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If we use Stirling's approximation, $10^8! \approx (10^8)^{(10^8)}e^{-10^8}=10^{8\cdot10^8}e^{-10^8}\lt 10^{10^9}$ by a factor of about $(100e)^{10^8}$ For numbers of this size, $\sqrt {2\pi10^8}$ is negligible.

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