  According to a well-known result in geometric topology, we have $\left
(\mathbb{S}^2 \right)^{n}\!\!/\operatorname{Sym}(n) = \mathbb{CP}^{n}$, where
$\operatorname{Sym}(n)$ acts on $\left (\mathbb{S}^2 \right)^{n}$ by coordinate
permutation. We use this fact to explicitly construct a regular simplicial cell
decomposition of $\mathbb{CP}^{n}$ for each $n \geq 2$. In more detail, we take
the standard two triangle crystallisation $S^2_3$ of the $2$-sphere
$\mathbb{S}^2$, in its $n$-fold Cartesian product. We then simplicially
subdivide, and prove that naively taking the $\operatorname{Sym}(n)$ quotient
yields a simplicial cell decomposition of $\mathbb{CP}^n$. Taking the first
derived subdivision of this cell complex produces a triangulation of
$\mathbb{CP}^n$. To the best of our knowledge, this is the first explicit
description of triangulations of $\mathbb{CP}^n$ for $n \geq 4$.
