Whilst going through online forums, I found this thread post, which claims
For $n\in Z >1$, $$S\cdot n^{\phi(r)} \leq \Phi_r(n)\leq \frac{1}{S}\cdot n^{\phi(r)}$$ where $ S = \prod_{m=1}^{\infty} (1-n^{-m})$
Apparently it should be easy to prove these bounds, but I am really struggling here. I simply can not make any progress.
I do know it has something to do with the following definition:
$$\Phi_r(n) = \prod_{d | r} (n^d-1)^{\mu(\frac{r}{d})} $$
Regardless I do not know how to approach this. I have never seen a similar result in any textbook or wikipedia. I would be thankful if someone could prove/disprove this.
Repeated from comment: Since $\deg \Phi_r = \phi(r)$, $$\Phi_r(n) / n^{\phi(r)} = \prod_{d \mid r} (1 - n^{-d})^{\mu(r/d)}.$$ Each of the factors $1 - n^{-d}$ is between $0$ and $1$, while $\mu(r/d) \in \{+1, -1, 0\}$. You get an upper bound just by replacing all the $\mu$'s with $-1$'s, and a lower bound by replacing all the $\mu$'s with $+1$'s. You get even worse upper and lower bounds by ignoring the restriction $d \mid r$.