The neutral pion belongs to the pseudoscalar meson octet, so it has, in the ground state ($L=0$):
\begin{align} P_{\pi^0}&=-1 \\ C_{\pi^0}&=+1. \end{align}
And the photon has
\begin{align} P_\gamma = -1 \\ C_\gamma=-1. \end{align}
Therefore, since electromagnetic interactions conserve parity and charge conjugation, why does the process
\begin{equation} \pi^0 \rightarrow \gamma\gamma \end{equation} occur? Doesn't it violate parity?
In the example I have seen in class, $C$ conservation is used to explain why the $\pi^0$ cannot decay into three photons, since for $\pi^0 \rightarrow \gamma\gamma\gamma$ we have
\begin{equation} C_i = +1 \neq C_f = (-1)^3 = -1 \end{equation}
and, for $\pi^0 \rightarrow \gamma\gamma$,
\begin{equation} C_i = +1 = C_f = (-1)^2 = +1, \end{equation} so regarding $C$ conservation it should be allowed. But, considering P conservation,
\begin{align} \pi^0 \rightarrow \gamma\gamma \qquad \Rightarrow \qquad P_i = (-1)^{L}\times \underbrace{(-1)}_{\text{intrinsic parity}} = -1 \neq P_f = (-1)^2 = +1 \end{align} so it would be forbidden for $L=0$. And, with the same argument, the decay into three photons would be allowed.
What am I missing?