I have seen a proposition: $\zeta_{p^2}$ is a root of $x^{p(p-1)}+x^{p(p-2)}+...+x^{2p}+x^p$, and $x^{p(p-1)}+x^{p(p-2)}+...+x^{2p}+x^p+1$ is the mininal polynomial of $\zeta_{p^2}$ over $\Bbb Q$.
But so far, by a simple substitution, I am not able to find a way to rearrange the equation and get zero, so could someone please help? May I please ask for a step-by-step deduction which leads to $0$? Thanks a lot.
We have to compute $\Phi_{p^2}(X)$. We have $\zeta_{p^2}$ is a root of $\Phi_{p^2}(X)$ by definition. Now we have $X^{p^2}-1$=$\Phi_{p^2}(X)$$\Phi_{p}(X)$$\Phi_{1}(X)$. We know $\Phi_{1}(X)=X-1$ and $\Phi_{p}(X)=\frac{(X^p-1)}{X-1}$. Thus we get $\Phi_{p^2}(X)=\frac{X^{p^2}-1}{X^p-1}$ which is exactly the polynomial you wrote.
we know $\zeta_{p^2}=e^{\frac{2\pi i}{p^2}}$. So $f(X)=\frac{X^{p^2}-1}{X^p-1}$ we have $f(\zeta_{p^2})=\frac{e^{2\pi i}-1}{e^{\frac{2\pi i}{p}}-1}=0$