approximation of binomial coefficient by exponentiation

Question

Show that inequality holds for every n, k. $$\ { n \choose k} \leq \frac{n^k}{k!}. ( \, 1-\frac {k}{2n} ) \,^{k-1} $$

I used Stirling's formula but I stucked, which is $$\ { n \choose k} = \frac{n^{n}}{(n-k)^{n-k}.k^k}. \sqrt \frac{n}{2\pi(n-k)k } $$

Then I distributed Binomial and stuck again in $$\ { n \choose k} = \prod_{i=1}^k \frac {n+1-i}{i} \leq \frac {n^k}{k!} $$

Answer 1

\begin{align} \binom nk&=\frac{n(n-1)\cdots(n-k+1)}{k!}\\ &=\frac n{k!}((n-1)(n-k+1))((n-2)(n-k+2))\cdots\\ &\le\frac n{k!}\left(n-\frac k2\right)^{k-1}\\ &=\frac{n^k}{k!}\left(1-\frac k{2n}\right)^{k-1}\;. \end{align}