Answer
True.
Work Step by Step
$f(x)=x^n;$ $f^{(n)}(x)=n!x^{0}$ which is a constant, then $f^{(n+1)}(x)=0.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 2 - Differentiation - 2.3 Exercises - Page 128: 132
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