How to prove which of two numbers written as powers is bigger?

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.