So the proof that the projection matrix has eigenvalues as only 1 or 0 is in this link . Then please explain why I can't extend the idea behind that proof as below: (forgive my lack of rigour, I'm only trying to get the idea across)
$\forall n\in \Bbb N (P^n=P)$(Because $P^n=P^{n-2}P^2=P^{n-2}P=P^{n-1}=..$ and so on until $..=P$)
Let λ an eigenvalue of P and x an eigenvector associated to it then
$\lambda x=Px=P^{n}x=P^{n-1}(Px)=P^{n-1}(\lambda^{} x)=P^{n-2}(\lambda Px)=P^{n-2}(\lambda^2x)=$.. and so on .. $=\lambda^nx$
So we get $\lambda=\lambda^n\implies\lambda^n-\lambda=0\implies\lambda(\lambda^{n-1}-1)=0\implies\lambda=0$ or $\lambda^n-1=0$
$\implies \lambda=0$ or $\lambda^{n-1}=1$
and we get $\lambda=0$ or $\lambda=exp\Bigl( i\frac{2\pi}{n-1}k\Bigr)$ for $k \in \{ 0,1,..,n-2\}$
If $n \in \mathbb N, n \ge 2$, then let $M_n:=\{\exp\Bigl( i\frac{2\pi}{n-1}k\Bigr):k \in \{ 0,1,..,n-2\}\}\cup \{0\}.$
You have shown: $\lambda$ is an eigenvalue of $P \implies \lambda \in M_n$.
But the reversed implikation is not true if $n \ge 3$.