I know that $\pi_1(CP^n)=0$ here is a possible proof:
Notation: for a CW complex, denote by $X^k$ the $k$-skeleton of $X$.
I will show that $\pi_1(CP^n)$ is contained in $\pi_1(S^2)=0$. Let $f:S^1 \longrightarrow CP^n$ be a map. By cellular approximation, $f$ can be assumed to be cellular. It follows that the image of $f$ lies in the $1$-skeleton of $CP^n$ that is: $f(S^1)\subset(CP^n)^1=S^2$. The statement follows.
Is this correct? Is there a more elementary way to show the result (cellular approximation isn't so trivial ) ? I think the long exact sequence of the fibration $S^1 \longrightarrow S^{2n+1} \longrightarrow CP^n $ only allows one to compute easily the $\pi_i$ for $i>2$.