February, 2017

Solution:
Let $V$, $E$, and $F$ denote the number of vertices, edges, and faces respectively. Let $a_k$ denote the number of faces with $k$ sides, and let $M$ be the maximum number of sides any face has.

Suppose that $a_k \leq n$ for all $k$ and that $M > 8$. Note that each edge is part of exactly two faces, and each vertex is part of at least three faces. It follows that
$$\begin{aligned} & \sum_{k=3}^{M} a_k = F \\ & \sum_{k=3}^{M} \frac{k a_k}{2} = E \\ & \sum_{k=3}^{M} \frac{k a_k}{3} \geq V \end{aligned}$$
and in particular
$$\sum_{k=3}^{M} a_k \left(1 - \frac{k}{2} + \frac{k}{3}\right) \geq F - E + V = 2$$
by Euler's formula. But on the other hand, $a_k \leq n$ by assumption, so
$$\begin{aligned} 2 &\leq \sum_{k=3}^{M} a_k \left(1 - \frac{k}{6}\right) \\ &\leq \sum_{k=3}^{M} n \left(1 - \frac{k}{6}\right) \\ &= \sum_{k=3}^{8} n \left(1 - \frac{k}{6}\right) + \sum_{k=9}^{M} n \left(1 - \frac{k}{6}\right) \\ &\leq \frac{1}{2} n - \frac{1}{2} n (M - 8) \end{aligned}$$
where the last step follows from the fact that $1 - \frac{k}{6} \leq -\frac{1}{2}$ for $k \geq 9$. Thus
$$2 \leq \frac{9}{2} n - \frac{1}{2} n M \Longrightarrow M \leq \frac{\frac{9}{2} n - 2}{\frac{1}{2} n} < 9$$
contradicting the fact that $M > 8$. It follows that $M \leq 8$, and as each face has at least 3 edges, the result follows directly from Pigeonhole.