Attempting to Prove a lower bound for the Binomial Coefficient The inequality I am having a problem with is as follows:
Let $1\leq k\leq n$, where $n,k$ are natural numbers. Prove that $\binom{n}{k}\geq(\frac{n}{k})^k$
I have proven already the inequality $\binom{n}{k}\leq\frac{n^k}{k!}\leq(\frac{en}{k})^k$, but cannot seem to prove my lower bound. I have been told that differentiating may be of use, though I am unsure how to approach this. Any help is greatly appreciated.

The ratio of $\binom{n}{k}$ and $\left(\frac nk\right)^k$ can be written as
$$\begin{align}
\frac{\binom{n}{k}}{\left(\frac nk\right)^k}&=\frac{\frac{n!}{n^k\,(n-k)!}}{k!/k^k}\\\\
&=\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)}{\left(1-\frac{1}{k}\right)\left(1-\frac{2}{k}\right)\left(1-\frac{3}{k}\right)\cdots\left(1-\frac{k-1}{k}\right)}
\end{align}$$
Inasmuch as $n\ge k$, each term in the numerator $\left(1-\frac{m}{n}\right)$ is greater than the corresponding term $\left(1-\frac{m}{k}\right)$  in the denominator.  Therefore, the ratio is bounded below by $1$ and we are done!
