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 ?

Solution 1:

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 $e$.

Solution 2:

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.