Without using a calculator and logarithm, which of $100^{101} , 101^{100}$ is greater? Which of the following numbers is greater? Without using a calculator and logarithm.
$$100^{101} , 101^{100}$$
My try :  $$100=10^2\\101=(100+1)=(10^2+1)$$
So  :
$$100^{101}=10^{2(101)}\\101^{100}=(10^2+1)^{100}=10^{2(100)}+N$$
Now what ?
 A: In this answer, it is shown that $\left(1+\frac1n\right)^n$ increases to $e\lt3$. Thus, for $n\ge3$,
$$
\left(1+\frac1n\right)^n\lt n
$$
Multiply by $n^n$ and we get
$$
(n+1)^n\lt n^{n+1}
$$
A: You want to determine if $\left(\frac{101}{100}\right)^{100}\geq 100$. But we know that $ \left(1+\frac{1}{n}\right)^n$ is always less than $e$.
A: 101100 = (100+1)100
Using the binomial theorem, we get 101 terms, none of them greater than 100100, and the three smallest of these terms add up to less than 100100. So the sum is less than 99(100100), which is less than 100(100100), which is equal to  100101.
A: $$\frac{101^{100}}{100^{101}} =\frac1{100}\cdot \left(\frac{101}{100}\right)^{100}=\frac1{100}\cdot\left(1+\frac1{100}\right)^{100}\approx \frac e{100}\ll 1$$
Of course, the "$\approx$" needs to be tamed, but the "$\ll$" allows for much leeway. One way to elaborate the idea is to use 
$$ \left(1+\frac1{100}\right)^{100}\cdot \left(1-\frac1{100}\right)^{100}=\left(1-\frac1{10000}\right)^{100}<1$$
and Bernoulli's inequality: 
$$\left(1-\frac1{100}\right)^{100}=\left(1-\frac1{100}\right)^{50}\cdot \left(1-\frac1{100}\right)^{50}\ge \left(1-\frac{50}{100}\right)^2=\frac14$$
so that ultimately 
$$\frac{101^{100}}{100^{101}}<\frac 4{100}. $$
A: for a sequence, $100^n$, as n increases, the number of digits increases by 2.
In the sequence of $101^n$, after finding the first 3 terms one can see the pattern and conclude that for all values of n, the term is 2 digits more than the last term, never 3 digits more. $101, 10201, 1030301, 104040401, 10505050501\dotsc$
If we compare $100^{101}$ and $101^{100}$, we can use the two above observations to conclude that $100^n > 101^{n-1}$, so $100^{101} > 101^{100}$.
