Proof for the upper bound and lower bound for binomial coefficients.
Solution 1:
First, we see that for $k=1$ we have a trivial statement $$\left( \frac{n}{k} \right)^k = n \leq n = \binom{n}{k}.$$
Now, consider $k > 1$ and let $0 < m < k \leq n$. Then
$$k \leq n \Rightarrow \frac{m}{n} \leq \frac{m}{k} \Rightarrow 1 - \frac{m}{k} \leq 1- \frac{m}{n} \Rightarrow \frac{k-m}{k} \leq \frac{n-m}{n} \Rightarrow \frac{n}{k} \leq \frac{n-m}{k-m}$$ Hence, $$\left( \frac{n}{k} \right)^k = \frac{n}{k} \cdot \ldots \cdot \frac{n}{k} \leq \frac{n}{k} \cdot \frac{n-1}{k-1} \cdot \ldots \cdot \frac{n-k+1}{1} = \binom{n}{k}.$$