Prove the estimate $\binom{n}{k} \le (\frac{en}{k})^k$

Question

Prove the estimate $$\binom{n}{k} \le \left(\frac{en}{k}\right)^k$$ directly from $$e\left(\frac{n}{e}\right)^n \le n! \le en\left(\frac{n}{e}\right)^n.$$

Answer 1

Use the other form of the binomial, after cancelling out $(n-k)!$ $$ \binom nk=\frac{n(n-1)\cdots(n-k+1)}{k!}\le \frac{n^k}{e(\frac{k}e)^k}=\frac1e\cdot\left(\frac{en}k\right)^k $$

Answer 2

Using Sterling's approximation(and that the quantities are asymptotic)

$\frac{n!}{k!(n-k)!} < \frac{n(n-1)...(n-k+1)}{\sqrt{2πk}*(k/e)^k}< (e/k)^k*(n*n*n*...*n) = (\frac{ne}{k})^k$