Prove that $\sum_{r=0}^k {{n}\choose{r}} < \frac{n^n}{k^k (n-k)^{n-k}}$

Question

Let $n\ge1$ and $k<\lfloor\frac{n}{2}\rfloor$
Prove that:
$$\sum_{r=0}^k {{n}\choose{r}} < \frac{n^n}{k^k (n-k)^{n-k}}$$

I know that (obvious):
$${{n}\choose{k}} < \frac{n^n}{k^k (n-k)^{n-k}}$$
and (can be proved by induction)
$$\sum_{r=0}^k {{n}\choose{r}} < \frac{n-k}{n-2k}{{n}\choose{k}}$$
But I don't understand how I can use this facts in order to prove first inequality.

Answer 1

$$\sum_{r=0}^k {{n}\choose{r}} = 1+ \sum_{r=1}^k {{n}\choose{r}} \le k{{n}\choose{k}} < {{n}\choose{k}}^k = \frac{n^k}{k^k (n-k)^{k}} < \frac{n^n}{k^k (n-k)^{n-k}}$$