Show that for any $n \ge 0$ and $0 \le k \le m$, the expression $\binom{n}{k}$ achieves its maximum value when $k = \lfloor n/2 \rfloor$ or $k = \lceil n/2 \rceil$.

It's evident in Pascal's triangle, yet anyway. There are $k$ multipliers in the denominator of:

$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

It's a question of minizing them. For $n/2 - k < i < n/2$, there would be $i$ pairs of the type $i(n-i)$. Those pairs are strictly greater than $i^2$. They are minized when $i$ is $n/2$ or the nearest integer.

It's not bullet-proof, but it works.