Covering the plane with convex polygons?

I have got the following task here:

Prove, that you can't cover the "Plane" with convex polygons, which have more than 6 vertices!

The answer is pretty obvious for $\phantom{\rule{thinmathspace}{0ex}}n=3\phantom{\rule{thinmathspace}{0ex}}$ vertices, because $6\cdot {60}^{\circ}={360}^{\circ}$.

For $\phantom{\rule{thinmathspace}{0ex}}n=4\phantom{\rule{thinmathspace}{0ex}}$ it works too, because $4\cdot {90}^{\circ}={360}^{\circ}$.

I think that $\phantom{\rule{thinmathspace}{0ex}}n=6\phantom{\rule{thinmathspace}{0ex}}$ is good too, but how do I prove, that other than that, it isn't possible to do that?

I have got the following task here:

Prove, that you can't cover the "Plane" with convex polygons, which have more than 6 vertices!

The answer is pretty obvious for $\phantom{\rule{thinmathspace}{0ex}}n=3\phantom{\rule{thinmathspace}{0ex}}$ vertices, because $6\cdot {60}^{\circ}={360}^{\circ}$.

For $\phantom{\rule{thinmathspace}{0ex}}n=4\phantom{\rule{thinmathspace}{0ex}}$ it works too, because $4\cdot {90}^{\circ}={360}^{\circ}$.

I think that $\phantom{\rule{thinmathspace}{0ex}}n=6\phantom{\rule{thinmathspace}{0ex}}$ is good too, but how do I prove, that other than that, it isn't possible to do that?