How to find the ith number that has n factors, considering that n isn't prime?

Question

The brute force approach would be checking every possible number.
If it has n factors: x++.

until: x = i.

but I just learned that you could get i = 1 that has n factors by:

1. Getting the set S of prime factors e of n.
2. Arrange set S in descending order.
3. Subtract 1 from every element e of set S.
4. Put a prime number p as base, such that:
   p_x-1< p_x, and treat e_x as exponent, for each element of set S.
5. Multiply every element.

Now given i = 1, What is the possible approach to get the ith term?

If you could provide an answer that would work on prime numbers that would be great.