let $F(k)$ be the number of faces of a convex polyhedron with k edges
how can we prove that $F(k) > 1$ for some $k$?
I know Euler's Formula for Polyhedra: $V-E+F=2$, and $\sum k\,F(k) = 2E$.
this means that some pair of faces has same number of edges for any polyhedron
here is restatement through dual: i want to prove that there is a pair of vertex with same number of edges.
Suppose not. Each face has a distinct number of edges.
Consider the face with the most number of edges, say $k$.
On each edge, there must be at least 1 distinct face, with at most $k-1$ edges.
We have $k$ faces, each has a distinct number of edges from 0 to $k-1$.
This is absurd. Hence contradiction.