Does $g = x^m - 1 \mid x^{mk} - 1$ for any $k \in \mathbb{N}$?

Question

Let $g = x^m - 1 \in \mathbb{Z}[x]$.

Question: Does $g = x^m - 1 \mid x^{mk} - 1$ for any $k \in \mathbb{N}$?

I know of only how to show the much weaker claim that $x^m - 1 \mid x^{m \cdot 2^k} - 1$ for any $k \in \mathbb{N}$.

Proof of Weaker Claim:

$(x^m - 1)(x^m + 1) = x^{2m} - 1$

$(x^{2m} - 1)(x^{2m} + 1) = x^{4m} - 1$

$\vdots$

$(x^{m \cdot 2^{k-1}} - 1)(x^{m \cdot 2^{k-1}} - 1) = x^{m \cdot 2^k} - 1$

But now what about the stronger claim that $x^m - 1 \mid x^{mk} - 1$ for any $k \in \mathbb{N}$?

Answer 1

Substitute $x^m = y.$ Then you are asking whether $y-1$ divides $y^k-1.$ Since the latter polynomial vanishes at $1,$ the answer is yes, but in fact, long division is your friend and the quotient is $\sum_{i=0}^{k-1} y^i.$

Answer 2

Yes, this is true. For a polynomial $p(x)$ with distinct roots to divide a polynomial $q(x)$, every root of $p$ must be a root of $q$.

By the derivative test, $x^m-1$ has distinct roots. Let $r$ be a root. Then $r^{mk}=(r^m)^k=1^k=1$, so $r$ is a root of $x^{mk-1}-1$. This proves the claim.