Example of a heptagonal polyhedron?

$$ V - E + F = 2 $$ let's see, $$ E = 7 F / 2 $$ Each vertex meets at least three faces, $$ V \leq 7F / 3. $$ $$ V - E + F \leq \frac{7F}{3} - \frac{5F}{2} = \frac{-F}{6} $$ $$ V - E + F \leq \frac{-F}{6} $$ $$ 2 \leq \frac{-F}{6} $$ which is bad

The Euler characteristic of a polyhedron $F + V - E = 2$

If we glue $n$ heptagons together we have

$F = n$

Since two faces meet at each edge

$E = \frac {7n}{2}$

And we must have at least 3 faces meeting at a vertex (unless you want to include degenerate heptagons with straight angles, and are really something with fewer sides)

$V \le \frac {7n}{3}$

and for any $n$

$F+V - E < 0$

You might be able to make some sort of torus, though.

Alternatively,

At each vertex the sum of the angles must be less than $360^\circ$ if the shape is convex.

And if we look at the differential between the sum of the angles and $360$ and sum it across all of the vertices, the sum equals $720^\circ$ if the surface is closed (and simply connected).

The average angle in a heptagon is $\frac {5}{7} 180^\circ$

The average vertex is concave. i.e. $ (1+ \frac {1}{7}) 360^\circ$

If you tile with heptagons you will either get "swiss cheese" i.e. a multi-holed torus, or you will get a model of a hyperbolic plane.