You would think that 99^101 < 101^99 just because you have a greater base. But the exponents in this case are the bigger telltale. First I'd use simpler powers of the bases. I'll go up to 5 for each of them.

101^1 -- self explanatory.

101^2 -- 10,201 (5 digits).

101^3 -- 1,030,301 (7 digits).

101^4 -- 104,060,401 (9 digits).

101^5 -- 10,510,100,501 (11 digits).

99^1 -- self explanatory.

99^2 -- 9,801 (4 digits).

99^3 -- 970,299 (6 digits).

99^4 -- 96,059,601 (8 digits).

99^5 -- 9,509,900,499 (10 digits).

See the patterns? 101^x will have 2x + 1 digits if x is a whole number at least 1, so 101^99 will have 2(99) + 1 = 199 digits. 99^x will have 2x digits with the same restriction for x, so 99^99 will only have 2(99) = 198 digits and won't surpass 101^99. But 99^100 will -- it would have 200 digits, and 99^101 would have 202 digits. The verdict:

99^101 > 101^99.