For $p$ prime, show that $\Phi_{p^n}(x) = 1 + x^{p^{n-1}} + x^{2p^{n-2}} + \dots +x^{(p-1)^{p^n-1}}$

Question

Well I attempted to try this but I failed to solve it: So $$\Phi_{p^{n}}(x)= \frac{x^{p^{n}} - 1}{\Phi_{1}(x) \Phi_{p^{2}}(x) \dots \Phi_{p^{n-1}}(x)}$$

Now I'm just stuck here. I saw a result somewhere on the net that said $\Phi_{p^n}(x) = \Phi_p(x^{p^{n-1}})$. I would just use that, but I don't see why that result is true or know how to show that's true.

How do I approach this?

Answer 1

See that $\Phi_{p^r}(X)\neq \Phi_{p}(X^{p^r-1})$ but $\Phi_{p^r}(X)= \Phi_{p}(X^{p^{r-1}})$

See that $$\Phi_{p^r}(X)=\frac{X^{p^r}-1}{\Phi_1(X)\Phi_p(X)\cdots\Phi_{p^{r-1}}(X)}=\frac{\left(X^{p^{r-1}}\right)^p-1}{X^{p^{r-1}}-1}=\Phi_p(X^{p^{r-1}})$$

Can you now apply this to your question?

Answer 2

We know that $x^n-1=\prod_{d\mid n}\Phi_d(x)$. Now let $n=p^k$ and use induction on $k$.