Can Someone pls explain the solution.

In order to understand the problem, we first need to analyse one key trick, the trick is

Given two numbers suppose "a" and "b" such that, b > a;
then we can conclude that, a^b > b^a, exceptions for the numbers 0,1,2,3;

Now, we can see that,

99^100 > 100^99

Now looking at the first part, for each quantity, we get,

10^(-100) and 9^(-99)
= 1/10^100 and 1/9^99

The above terms are so small infront of the their second term, that we can completely neglect them.

Hence the final question becomes, to compare between 99^100 and 100^99

And clearly 99^100 > 100^99

So Answer is option A