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 ?

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.
