Short and intuitive proof that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$ The simple inequality that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$ has a number of different proofs. But is there a particularly intuitive, short and elegant proof that uses the natural interpretation of binomial coefficients, for example. I would ideally like a proof which is also accessible to students with very limited prerequisite knowledge.
Here is the best proof that I have seen which is less intuitive than I was hoping for.
First we first prove that
$$\frac{n-i}{k-i} \geq \frac{n}{k}$$
 for $i<k\leq n$.  This follows from 
$$0\leq (n-k)i = k(n-i) - n(k-i) = 
k(k-i)\left(\frac{n-i}{k-i}-\frac{n}{k}\right),$$
and $k(k-i)> 0$, so $(n-i)/(k-i) \geq n/k.$
Now we multiply the over
$i\in\{1,\ldots,k-1\}$ to obtain 
$$\frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots 1} \geq \frac{n^k}{k^k},$$
or equivalently $\binom{n}{k}\geq (n/k)^k$.
 A: Note first that for all $0 \leq i < k \leq n$ we have $k(n-i) = kn - ki > kn - ni =\geq n(k-i)$, and hence $\frac{n-i}{k-i} \geq \frac{n}{k}$.
Therefore
$${n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)}{k \cdot (k-1) \cdots 1} = \prod_{i=0}^{k-1} \frac{n-i}{k-i} \geq \prod_{i=0}^{k-1} \frac{n}{k} = \left(\frac{n}{k}\right)^k.$$
A: For every  $1\leqslant k\leqslant n$ and $0\leqslant i\leqslant k-1$,
$$
k\leqslant n\implies\frac{i}k\geqslant\frac{i}n\implies 1-\frac{i}k\leqslant1-\frac{i}n.
$$
Each term is positive, hence the products are in the same order, that is,
$$
\frac{k!}{k^k}=\prod_{i=0}^{k-1}\left(1-\frac{i}k\right)\leqslant\prod_{i=0}^{k-1}\left(1-\frac{i}n\right)=\frac{n!}{n^k(n-k)!}.
$$
Multiply the leftmost and rightmost terms by $\dfrac{n^k}{k!}$... Et voilà!
A: $n^k$ is the number of ways of picking $k$ balls from $n$ balls with repetition allowed. One can generate all the possible ways by first deciding which $k$ out of $n$ balls to draw and
draw from the $k$ selected balls instead. There are $\binom{n}{k}$ ways to choose the $k$ balls and $k^k$ ways to pick from the selected $k$ balls with repetition allowed. This gives us
$$n^k \le\binom{n}{k} k^k \quad\iff\quad \left(\frac{n}{k}\right)^k \le \binom{n}{k}$$
