How to prove which of two numbers written as powers is bigger? Prove which number is larger:
a) $10^{100!}$ or $10^{10^{100}}$
b) $e^\pi$ or $\pi^e$
I know we all know how to plug these into the calculator and check, but how someone mathematically prove which one is bigger with words and calculations? 
 A: For (a), it boils to comparing $1$ and
$$
\frac{100!}{10^{100}}=\frac{1}{10}\cdot\frac{2}{10}\cdot\cdots\frac{100}{10}.\tag{*}
$$
Only the fractions $\frac{1}{10},\ldots,\frac{9}{10}$ are less than $1$ but they are more than compensated for by $\frac{100}{10},\frac{90}{10},\ldots,\frac{20}{10}$. Other fractions are greater or equal to $1$. So it should be clear that the expression in (*) is greater than $1$. This of course means $10^{100!}>10^{10^{100}}$.
For (b), an answer is already linked to in the comment or you can look at Edward Jiang's answer.
A: For $a)$, you are comparing 
$$
10^{100} \text{ and } 10^{10^{100}} 
$$
Realize that:
$$
10^{10^{100}} = 10^{({10^{2}})^{50}}=(10^{100^{50}}) =10^{100\cdot 100\cdot100\cdot\ldots\cdot 100}=((10^{100})^{100})^{100} )^{100}\ldots)^{100} 
$$
Use the fact that for the function $f(x) = x^{100}$, $f(x) > x$ for all $x > 1$. 
This can be easily seen as let:
$$
g(x) = f(x) - x
$$
Then $$
g'(x) = 100x^{99} -1 >0
$$
for all $x>1$
This gives $10^{100}>10$
$$(10^{100})^{100}>10^{100}\\ \vdots \\ ((10^{100})^{100})^{100} )^{100}\underbrace{\ldots}_{50})^{100} > ((10^{100})^{100})^{100} )^{100}\underbrace{\ldots}_{49})^{100} $$, etc. 
For b) see mrf's advice in the comments
A: For (a) notice that $f(x) = 10^x$ is a strictly increasing function and $2<100 \Rightarrow 10^2<10^{100} \Rightarrow 10^{10^2} < 10^{10^{100}}$.
For (b) see in the comments. 
