Answer:
The simpler case is the number:
n = a^7, where a is prime.
We use this so we avoid having cross numbers by multipliying different prime numbers. because for example lets compare:
2^2 = 4 has 3 divisors, 4, 2 and 1.
(4*4) = 16 has the divisors: 1, 2, 4, 8 and 16. (5 divisors).
Now let's mix primes:
2*3 = 6 has 3 divisors, 3, 2 and 1.
6*6 = 36
(3*2*1)*(3*2*1) = 36
then the divisors are:
3*3, 3*2, 3*1, 2*2, 2*1, 1*1, 3*2*2, ..., etc.
You can clearly see that this has a lot more divisors than the previous case.
Then returning to the answer:
a^7 can be divided by a, a^2, a^3, ..., and by itself, a^7 (and 1, so we have exactly 8 divisors).
Now, we can write:
n^2 = (a^7)^2 = (a^(2*7)) = a^14.
By the same logic that was used above, a^14 has 14 + 1 divisors.
Then the smallest number of divisors possible is 15.
