Proof that one large number is larger than another large number Let $a = (10^n - 1)^{(10^n)}$ and $b=(10^n)^{(10^n - 1)}$
Which of these numbers is greater as n gets large?
I believe it is $a$ after looking at some smaller special cases, but I'm not sure how to prove it.
 A: Divide:
$$\frac{a}{b} = \frac{(10^n-1)^{10^n}}{(10^n)^{10^n-1}} = (10^n-1)\left(1- \frac{1}{10^{n}}\right)^{10^n-1} \approx (10^n-1)e^{-1} > 1.$$
A: Because the logarithm is monotonic, you can take the difference of the logs:  $$\log a - \log b=10^n \log(10^n-1)-(10^n-1)\log 10^n\\
=10^n(\log(10^n)+\log(1-\frac 1{10^n})-\left(n(10^n-1)\log 10\right)
\\ \approx n10^n\log(10)-1-n10^n\log(10)+n\log(10)\\ =n\log(10)-1 \gt 0$$
A: Say $x={ 10 }^{ n }\\$ $$\frac { { \left( x-1 \right)  }^{ x } }{ { x }^{ x-1 } } =x{ \left( \frac { x-1 }{ x }  \right)  }^{ x }=x{ \left( 1-\frac { 1 }{ x }  \right)  }^{ x }\\ \lim _{ x\rightarrow \infty  }{ x{ \left( 1-\frac { 1 }{ x }  \right)  }^{ x } } =\lim _{ x\rightarrow \infty  }{ \frac { x }{ e }  } =\infty $$
According to calculation on Wolframalpha for any $x > 3.2932$, $$(x-1)^{ x }>x^{x-1}$$
A: Let $x=10^n$, $y=10^n-1$.
The function $\frac {x} {\ln x}$ is increasing from $e$.
Both $x$ and $y$ are greater than $e$, so $\frac {x} {\ln x} > \frac {y} {\ln y} \Rightarrow x\ln y > y\ln x \Rightarrow 
y^x > x^y$
