Let $a = 2019^{1009} , b = 2019!$ and $c = 1010^{2019}$

Question

Let $a = 2019^{1009} , b = 2019!$ and $c = 1010^{2019}$. Arrange $a,b,c$ in ascending order.

What I Tried:- I have solved one of the parts.

We have $2019 < 1010^2$.
$\implies 2019^{1009} < ({1010^2})^{1009} = 1010^{2018} < 1010^{2019}.$

Now, I am stuck on how to compare $a,c$ with $2019!$. I tried it for $2$ hrs, but I still did not get any idea.

Can someone help me? Thank You.

Note :- I might have seen this comparison of $2019!$ before but my searching did not give me results, sorry about that.

Answer 1

Here's a hint: consider what $2019!$ means, and then how can you rewrite the product to be comparable?

The following spoiler sort of gives it away:

$2019!=2019\times2018\times2017\times\dots\times2\times1=(2019\times1)(2018\times2)\dots(1011\times1009)(1010)$

And a full solution for completeness:

$2019!$ consists of the product of $1009$ numbers that are all larger than or equal to $2019$, multiplied by $1010$, and therefore $\frac{2019!}{1010}>2019^{1009}$ so $2019!>2019^{1009}$. It is also the product of $1009$ numbers that are all smaller than or equal to $1010^2$, times $1010$ and therefore $2019!<1010\times(1010^2)^{1009}=1010^{2019}$.

Here's a quick Python script to show that this result is correct. Of course not a proof, but mostly just a sanity check for myself to make sure I did it correctly, honestly.