Properties of cyclotomic polynomial

Question

Assume first that $p$ a prime divides $n$. I have to show that $\Phi_{np}(X)=\Phi_n(X^p)$. Here is what I tried: Suppose $\eta_i$ are roots of $\Phi_{np}(X)$ so $\eta_i=\text{exp}(\frac{2\pi i a_i}{np})$ with $(a_i,np)=1$, but then also $(a_i,n)=1$. Now fill in $\eta_i$ in $\Phi_{n}(X^p)$ and we see $$\Phi_{n}(\eta_i^p)=\Phi_{n}(e^{2\pi i a_i/n})=0$$ because $\eta_i^p$ is a primitive nth-root. We can conclude $\Phi_{np}(X)$ divides $\Phi_{n}(X^p)$. However, I can not show that they are equal. Because I don't understand how they could be equal. Degree of $\Phi_{np}(X)=\phi(np)<\phi(n)p$ and the degree of $\Phi_n(X^p)=\phi(n)p$, or is this wrong?