No simple polyhedron has seven edges.

Proof: We show first that for any polyhedron we have 2E ≥ 3F and 2E ≥ 3V. The faces of the polyhedron are polygons, each bounded by a number of sides. Along each edge exactly two faces come together, so an edge corresponds to exactly two sides: the total number of sides is 2E. We also notice that any face has at least 3 sides, so the total number of sides is at least 3 times the number of faces. Thus we get:

The total number of sides = 2E

and

The total number of sides ≥ 3F.

Putting this together we get:

2E ≥ 3F,

proving our first inequality.

To prove the second inequality we count the total number of ends of edges. Each edge has two ends, so the total number of ends equals 2E. At each vertex at least three edges come together, so the total number of ends of edges is at least 3 times the number of vertices. Putting this together we get:

2E ≥ 3V.

Now, if a polyhedron has 7 edges, then 3F ≤ 14 and 3V ≤ 14. This means that both F and V cannot be bigger than 4. A little thought will convince you that every polyhedron has strictly more than three faces, so we must have F=4. Similarly we get that V=4. This gives

V - E + F = 4 - 7 + 4 = 1 ≠ 2.

This tells us that our hypothetical seven-edged polyhedron cannot exist, for if it did, then Euler's formula would hold and the above sum would have to be equal to 2 — QED!