Is $10^{8}!$ greater than $10^{10^9}$? 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,
 A: 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 $10^8!<10^{10^9}$.
A: $10^8!=\prod \limits_{n=1}^{10^8}n<\prod \limits_{n=1}^{10^8}10^8=(10^8)^{10^8}=10^{8\times10^8}<10^{10\times10^8}=(10)^{10^9}$
A: 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.
A: 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 $10^8!<(10^8)^{10^8}=10^{8\cdot 10^8}<10^{10^9}$.
