How to prove that a hexagon is the regular polygon with the most sides that can tile a plane

The simplest way is to begin as you did, and consider how many polygons are meeting at a corner. It has to be at least 3 to be a corner; if it were 2 it would be an edge, not a vertex. The minimum is 3. If all the angles are the same size, then the size of each interior angle that joins at the vertex must be 360/3 = 120. Therefore a hexagon has the most sides.

Well, $\lim_{n \rightarrow \infty} \frac{2n}{n-2} = 2$. Its derivative, $\frac{-4}{(n-2)^2}$ is negative everywhere, so this expression decreases asymptotically to $2$. So $3$ is the least integer it can take and you have shown that it does.

which simplifies down to $\frac {2n}{n-2}$. Now, I need to prove that n=6 is the biggest number such that this quanitity is an integer

Well, $\frac {2n}{n-2} = \frac {2n -4 + 4}{n-2} = 2 + \frac 4{n-2}$.

That can only be an integer if $n-2|4$. Or in other words if $n-2 = 1,2$ or $4$ or if $n = 3, 4$ or $6$.

In particular if $n > 6$ then $4{n-2} < 1$ and can't be an integer.

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Alternatively $\frac {2n}{n-2} > \frac {2n}{n} =2$ so $3$ is the smallest possible integer option. And $\frac {2n}{n-2} -3$ occurs if $n = 6$.

If $n > 6$ then $\frac {2n}{n-2} < 3$