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We know that for a given product the sum of all the numbers will be the least when all the numbers are equal
Thus, the sum of all the number in $$n^n$$ will be least when
$$n^n$$ = $$n*n*n*...........*(n \text{times})$$
Thus, their sum $$= n+n+........+(n times) = n^2$$
Hence, the sum can never be less than $$n^2$$
Hence, option D is the correct answer.
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