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This is a claim in the book "Abstract algebra" in one of the examples, can someone explain this please ?

I know that the derivative of this polynomial is identically $0$ so there is a multiple root, but why all the roots are multiple ?

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    Denote $P = X^n-1$. If $P(\lambda) = 0$, then you also have $P'(\lambda) = 0$ (because $P' = 0$), so $\lambda$ is a root of order at least two.2012-05-04

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If $n=kp$ then $x^n -1= (x^k -1)^p$, because $x\mapsto x^p$ is an automorphism. So, the roots of $x^n -1$ are exactly the roots of $x^k -1$ and appear with multiplicity at least $p$.

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    I don't follow, can you please add more details ?2012-05-04
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Because $x^{dp}-1 = (x^d-1)^p$