If $p\neq 2$, then the cyclic group $\mathbb{Z}_p$ has no free continuous action on $\mathbb{C}P^n$. My question is how to prove the above fact using Leray-Serre spectral sequence associated to the Borel fibration $ \mathbb{C}P^n\hookrightarrow X_{\mathbb{Z}_p}\rightarrow B_{\mathbb{Z}_p}$.

From the Euler Characteristic argument, $p$ divides $n+1$. Also, $\pi_1(B_{\mathbb{Z}_p})={\mathbb{Z}_p}$ acts trivially on $H^*(\mathbb{C}P^n;\mathbb{Z}_p)$ by using Lefschetz fixed point theorem. $H^*(\mathbb{C}P^n;\mathbb{Z}_p)=\mathbb{Z}_p[b]/\langle b^{n+1}\rangle$ and $H^*(\mathbb{Z}_p;\mathbb{Z}_p)=\bigwedge(s)\otimes\mathbb{Z}_p[t]$. So the only possibility is $d_3(b)=st$. It follows $d_3(sb)=0$ and $d_3{(tb)}=st^2$. After that, I am unable to deduce any contradiction.

Thank you so much in advance.