How do we show that $(10^{10})!$ is larger than $10^{10^{10}}$? The question is simply: how do we show that $(10^{10})!$ is larger than $10^{10^{10}}$?
The context in which this question arose was in trying to find a polynomial approximation to $\sin{1}$ with an error of less than $10^{-10^{10}}$, the details of which are in this other question.
In summary, if one knows that $(10^{10})!$ is larger than $10^{10^{10}}$, then it turns out a Taylor Polynomial of order $10^{10}$ is sufficient to get an error lower than $10^{-10^{10}}$.
 A: $10^{10^{10}}$ is the product of $10^{10}$ factors of $10$ while the first $\frac 12 \cdot 10^{10}$ factors of $(10^{10})!$ are larger than $10^2$ so have a product larger than $10^{10^{10}}$.
A: $$\left(10^{10}\right)! = 100!\cdot\left(\prod\limits_{i=101}^{10^{10}}i\right) \gt 100!\cdot\left(\prod\limits_{i=101}^{10^{10}}100\right) = 100!\cdot 100^{10^{10}-100} = 100!\cdot 10^{2\cdot10^{10} - 200}$$ $$\gt 10^{2\cdot10^{10} - 200} = 10^{10^{10} + 10^{10} - 200} \gt 10^{10^{10}}$$
This uses the lemma that $10^{10} \gt 200$.
A: Using algebra, you want to find when
$$(a^a)! \geq a^{a^a}$$ To make life easier because of the huge numbers, when is
$$f(a)=\log\Bigg[\frac{(a^a)!}{a^{a^a}}\Bigg] >0$$
By inspection or plotting, this is already true if $a>2$ since $f(2)=\log \left(\frac{3}{2}\right)$.
More precisely, using Newton method, the zero of $f(a)$ is $1.83438$ which does not seem to be known by inverse symbolic calculators.
A: For integer valued $a$, $a! > a^{10}$ for $a\geq 21$; ($ a= 10^{10}\geq  21$) .
$$a! = a(a-1)(a-2)(a-3) .... (a-19)...1 > a*[(a-1)(a-19)] *[(a-2)(a-18)]*...*[(a-9)(a-10)] > a [(a-1)(a-19)]^9 \geq a(2a-2)^9 > a^{10}.$$
