I think the comments answer the question, but to give you a reference:

Milnor, Stasheff: <A href="http://books.google.de/books?id=5zQ9AFk1i4EC&printsec=frontcover&source=gbs_slider_thumb#v=onepage&q&f=false">Characteristic Classes</a>, Chapter 6

They prove that every Grasmann manifold $G_n(\mathbb{R}^m)$ is a CW-Complex. (The cells are constructed with Schubert symbols). The complex case works in the same fashion.<br> 
As a result you get that $\mathbb{CP}^n$ consists of $n+1$ cells: for every $0 \leq k \leq n$ you get one $2k$-cell. The $2k$-skeleton is a $\mathbb{CP}^k$ 

**EDIT:** Sorry for the sloppiness!<br>
Not every CW-Complex is triangulable, but the following holds:<br>
**Every regular CW-Complex (<strike>and $\mathbb{CP}^n$ is a regular complex</strike> $\oplus$) $X$ is triangulable**.<br> 
This is true, since the barycentric subdivision is a simplicial complex that is homeomorphic to $X$. For a full proof, see for example <a href="http://books.google.de/books?id=J-2F641zx-MC&pg=PA130&lpg=PA130&dq=CW-Complex+triangulable&source=bl&ots=857Bjqzpce&sig=RZO51f71tCOvbnqiyIKfSAO_isQ&hl=de&ei=bla-S4-lOYOvOPqK1ZYE&sa=X&oi=book_result&ct=result&resnum=2&ved=0CA4Q6AEwATgK#v=onepage&q=CW-Complex%20triangulable&f=false">Cellular structures in topology</a> (p.130) by Fritsch and Piccinini.

**Edit 2:** $\oplus$: Perhaps the next sloppiness: The CW-structure of $\mathbb{CP}^n$ obtained by Schubert cells isn't regular (the characteristic map is 2-to-1) but I think there exists a regular CW-structure. But this might be harder to prove than I thought?!