Let $f\in\mathbb{Q}[x]$ and suppose that $f(x)=q(x)\Phi_p(x)^k$, for some $k\geq 1$ with $q\in\mathbb{Q}[x]$ and $\gcd(q(x),\Phi_p(x))=1$, where $\Phi_p$ is cyclotomic polynomial with $p>2$ prime. Show that $f(x^2)=r(x)\Phi_p(x)^k$ for the same $k\geq 1$ and for some $r\in\mathbb{Q}[x]$ with $\gcd(r(x),\Phi_p(x))=1$.
I'm not sure how to go about this. This is my attempt so far: I've noticed that $$ \Phi_p(x^2)=\frac{x^{2p}-1}{x^2-1}=\frac{(x^p-1)(x^p+1)}{(x-1)(x+1)}=\Phi_p(x)\frac{x^p+1}{x+1}, $$ So $$ f(x^2)=q(x^2)\Phi_p(x^2)^k=q(x^2)\left(\frac{x^p+1}{x+1}\right)^k\Phi_p(x)^k. $$ Then for this to be true, it must be the case that $$ r(x)=q(x^2)\left(\frac{x^p+1}{x+1}\right)^k. $$ The problem is I'm pretty sure that $q(x^2)\left(\tfrac{x^p+1}{x+1}\right)^k$ is not a polynomial (let alone with coefficients in $\mathbb{Q}$), so I think I may be going about this wrong.
This is homework so only hints are appreciated. Thanks in advance.