Cyclotomic Polynomial Recurrence Identity 
If p is prime, show that $\Phi_{p}(x^{p^{k-1}})=\Phi_{p^{k}}(x)$.

I have the solution, however I cannot make sense of how the author got to it. I understand the need to use the fact that p^k and n are coprime if and only if p and n are coprime, however the rest of the details are vague to me. I apologise for my formatting if I messed it up. Thank you
http://people.maths.ox.ac.uk/earl/complex/0163.pdf  - Intended Solution (would appreciate it if someone could explain it)

The number of coprimes is given by the euler totient which says the number of coprimes is $\phi(p^n)=p^n-p^{n-1}$ whenever $p$ is prime so what is seems to me the proof is doing is something like
$\Phi_p(x)=\frac{x^p-1}{x-1}\Rightarrow \Phi_p(x^{p^{n-1}})=\frac{x^{p^n}-1}{x^{p^{n-1}}-1}$
$\Phi_{p^1}(x)=\prod_{k,\gcd(k,p^1)=1}(x-\omega_k)=\prod_{k=1}^{\phi(p)}(x-\omega_k)=\prod_{k=1}^{p-1}(x-\omega_k)=\frac{\prod_{k=1}^p(x-\omega_k)}{\prod_{k=1}^1(x-\omega_k)}=\frac{x^p-1}{x-1}=\Phi_p(x)$
$\Phi_{p^n}(x)=\prod_{k,\gcd(k,p^n)=1}(x-\omega_k)=\prod_{k=1}^{\phi(p^n)}(x-\omega_k)=\prod_{k=1}^{p^n-p^{n-1}}(x-\omega_k)=\frac{\prod_{k=1}^{p^n}(x-\omega_k)}{\prod_{k=1}^{p^{n-1}}(x-\omega_k)}=\frac{x^{p^n}-1}{x^{p^{n-1}}-1}=\Phi_p(x^{p^{n-1}})$

From exploring this problem I think the proof goes this way to avoid proving it as is appears to be more commonly done but which is maybe a little more complicated, something along the lines
Take/proof the identity
$\Phi_n(x)=\frac{x^n-1}{\prod_{d|n}\Phi_d(x)}$
Which for $n=p^n$ implies 
$\Phi_{p^{n}}(x)=\frac{x^{p^{n}}-1}{\prod_{k=0}^{n-1}\Phi_{p^{k}}(x)}\quad (d|p^n\Rightarrow d=p^k,\ k<n)$
and then proceed inductively
$\Phi_{p^{0}}(x)=x^{p^{0}}-1=x-1=\Phi_1(x)$
$\Phi_{p^1}(x)=\frac{x^{p^{1}}-1}{\prod_{k=0}^{0}\Phi_{p^{k}}(x)}=\frac{x^{p^{1}}-1}{\Phi_{p^{0}}(x)}=\frac{x^{p^{1}}-1}{x-1}=\Phi_{p}(x^{p^0})$
$\Phi_{p^2}(x)=\frac{x^{p^{2}}-1}{\prod_{k=0}^{1}\Phi_{p^{k}}(x)}=\frac{x^{p^{2}}-1}{\Phi_{p^{0}}(x)\Phi_{p^{1}}(x)}=\frac{x^{p^{2}}-1}{x^{p^{1}}-1}=\Phi_{p}(x^{p^1})$
and so on
