Prove that there is no convex polyhedron with exactly $7$ edges
Solution: We show first that for any polyhedron we have $2E \geq 3F$ and $2E \geq 3V$. 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 $\geq 3F$.
Putting this together we get:
$2E \geq 3F$,
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:
$2E \geq 3V$.
Now, if a polyhedron has $7$ edges, then $3F \leq 14$ and $3V \leq 14$. 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
$$V - E + F = 4 - 7 + 4 = 1 \neq 2$$.
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 $2$
Question: Are there any more formal or shorter demonstrations than this? If not, how to proceed with a different solution?
Notice that the Euler theorem is not required. Since you found $V=4$, the polyhedron is necessarily a tetrahedron(and this can be proved easily by exhaustion). This procedure is more general, because it removes the convexity hyphothesis.
There are no polyhedron with 7 edges.
Even more generally you can prove easily that there exist polyhedron with $n$ edges with n≥6, with 7 being the only exception.