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 f(x) = xⁿ 

f'(x) = n x(n-1) 

f''(x) = n(n-1) x(n-2) 

f'''(x) = n(n-1)(n-2) x(n-3) 

fⁿ(x) = n! x(n-n) , and since n-n = 0, x0 =1, 

so fⁿ(x) = n! 

Hence,Option(D)n! is the correct choice.
 

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