Prime number problem, pure maths, explain this solution

[smiddy]

Homework Statement

Prove that for every k >= 2 there exists a number with precisely k divisors.


I know the solution, but don't fully understand it, here it is;


Consider any prime p. Let n = p^(k-1). An integer divides n if and only if it has the form p^i where 0<= i <= (k-1). There are k choices for i, therefore n has exactly k divisors.

Could someone fully explain the thought process involved in finding the solution, I understand p^i etc, just don't know where p^(k-1) comes from.

 

[Office_Shredder]

If you understood the though process, you'd know where p^{k-1} came from. Given the problem, one can decide to be clever and take a prime, say p. Then if we look at

1,p,p²,...p^k-1

Then 1, p, p², ... ,p^k-1 all divide p^k-1, and no other numbers do. Hence, we found a number with exactly k divisors