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