Five Platonic Solids/Proof 2

Proof
Consider a convex regular polyhedron $P$.

Let $m$ be the number of sides of each of the regular polygons that form the faces of $P$.

Let $n$ be the number of those polygons which meet at each vertex of $P$.

From Internal Angles of Regular Polygon, the internal angles of each face of $P$ measure $180^\circ - \dfrac {360^\circ} m$.

The sum of the internal angles must be less than $360^\circ$.

So:

But $m$ and $n$ are both at greater than $2$.

So:
 * if $m = 3$, $n$ can only be $3$, $4$ or $5$
 * if $m = 4$, $n$ can only be $3$
 * if $m = 5$, $n$ can only be $3$

and $m$ cannot be greater than $3$.

There are $5$ possibilities in all.

Therefore all platonic solids have been accounted for.