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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.

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# Homework Help: Prime number problem, pure maths, explain this solution

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