Q: Cyclotomic polynomial identity proof with primes. 
If p is prime, show that $\Phi_{p}(x^{p^{k-1}})=\Phi_{p^{k}}(x)$.

Here is my attempt:
$x^{p^{k}}-1=\prod_{d|p^{k}}\Phi_{d}(x)=\prod_{i=1}^{k}\Phi_{p^{i}}(x)=\Phi_{p^{k}}(x)(\Phi_{p}\Phi_{p^{2}}\cdots \Phi_{p^{k-1}})=(x^{p^{k-1}})^{p}-1=(x^{p^{k-1}}-1)\Phi_{p}(x^{p^{k-1}})$
But how do I show that $x^{p^{k-1}}-1=\Phi_{p}\Phi_{p^{2}}\cdots \Phi_{p^{k-1}}$?
