for a sequence, 100^n$100^n$, as n increases, the number of digits increases by 2.
In the sequence of 101^n$101^n$, after finding the first 3 terms one can see the pattern and conclude that for all values of n, the term is 2 digits more than the last term, never 3 digits more. 101, 10201, 1030301, 104040401, 10505050501...$101, 10201, 1030301, 104040401, 10505050501\dotsc$
If we compare 100^101$100^{101}$ and 101^100$101^{100}$, we can use the two above observations to conclude that 100^n > 101^(n-1)$100^n > 101^{n-1}$, so 100^101 > 101^100$100^{101} > 101^{100}$.
