To begin, note that this is working out as a sort of a pattern, where (15)0=1,(15)1=15, and then it continues on to where the exponential power keeps increasing by 1. Let's express this in sigma notation:
∑∞n=0(15)n, because there is no ending to the original sum, so we assume that it goes on forever to infinity. Now, we evaluate the sum, using the formula for geometeric series: a(1−rn1−r), where we know a is the first term, which is 1. R is the common ratio, which we see that we are raising 1/5 to some power, so the common ratio is 1/5. Then, we have n, which is the number of terms. This is infinity, because the index of the summation ends at infinity. So, writing this out gives:
a(1−rn1−r)⟹1(1−(15)∞1−15).
When raising something to the infinite power, we conclude that it is 0, so for the numerator, we have 1 - 0, which is 1, and for the denominator, we have 1 - 1/5, which is 4/5. 1 divided by 4/5 is 1 times 5/4, so the answer is 54.
Please correct me if I made a mistake, I am always willing to learn! 