Recall the rules of #i#. #i# is defined such that:
#i = i#
#i^2 = -1#
#i^3 = i*i^2 = -i#
#i^4 = (i^2)^2 = (-1)^2 = 1#
We also know that exponent rules function such that #(x^a)/(x^b) = x^(a-b)#. If we divide #i^101# by some factor of #i^101#, such as #i^x#, where #0<=x<101# and where #x# is divisible by 4, we will yield #i^(101-x)#
With this in mind, return to the initial problem. What is the largest power of #i# we can factor out of #i^101# (e.g. what is the highest #x# divisible by 4 and less than 101? This should be obtainable by subtracting 1 from 101 as many times as it takes to yield a number divisible by 4. As it turns out, we reach that point at #x=100#.
From here, the calculation is simple.
#i^101 = (i^101)/1 = (i^101)/(i^100) = i^(101-100) = i^1 = i#
