The first move converts the polyhedron into a planar graph. Since the polyhedron is convex, its surface is homeomorphic to a sphere $S^2$; remove one face — think of it as the 'back' face — and the remaining surface is topologically a disk, which can be flattened into the plane. Concretely, place the polyhedron so the chosen face is closest to a point light source (or use stereographic projection from a point inside that face): the edges project to a planar embedding in which all other faces become bounded regions and the removed face becomes the outer, unbounded region.

The bookkeeping is immediate: vertices and edges are preserved one-to-one, so $V_{\text{graph}}=V$ and $E_{\text{graph}}=E$; the bounded faces of the drawing number $F-1$, and counting the outer face brings the total back to $F$. Hence the quantity $V-E+F$ is literally unchanged by the flattening — proving it for the planar graph proves it for the polyhedron.

This is the moment where the hypothesis 'convex' is actually used: convexity is what guarantees the surface is a sphere and the projection is a genuine embedding with no edge crossings. The proof is the one given in Aigner–Ziegler, Proofs from THE BOOK, Ch. 13.