Proof $\left|1-\binom{n}{k} \frac{k!}{n^k}\right| \leq \frac{k(k-1)}{n}$

Question

For all $n \in \mathbb{N}, 0 \leq k \leq n$ it holds that

$$\left|1-\binom{n}{k} \frac{k!}{n^k}\right| \leq \frac{k(k-1)}{n}$$

How can one prove that?

Base case: $$ \left|1-\frac{1!}{k!\cdot(1-k)!} \frac{k!}{1^k}\right| = \left|1-\frac{1!}{(1-k)!} \frac{1}{1^k}\right| $$

If $k=0$, then

$$\left|1-\frac{1!}{(1-1)!} \frac{0}{1^0}\right| = 1 - 1 * 0 = 0 = \frac{0(0-1)}{1} = 0$$

If $k=1$, then

$$1-\frac{1!}{(1-1)!}\frac{1}{1^1} = 0 = \frac{1(1-1)}{1} = 0$$

Induction step:

$$ 1- \binom{n+1}{k}\frac{k!}{(n+1)^k} = 1 - \frac{(n+1)!}{k! \cdot (n+1-k)!} \frac{k!}{(n+1)^k} = \frac{(n+1)!}{(n+1-k)! (n+1)^k} = ?$$

Answer 1

The proposed inequality rhymes well with the following general inequality which can be proved by induction

Let $I$ be a finite set (of indices) and $a \in [0,1]^I$ a family of numbers. It then holds that: $$\prod_{i \in I}(1-a_i) \geqslant 1- \sum_{i \in I} a_i$$

(needless to say, the induction is carried out with respect to the cardinal of $I$).

In your particular case you have $$\left|1-{n\choose k}\left( \frac{k!}{n^k} \right)\right|=\left|1-\prod_{h=0}^{k-1}\left(1-\frac{h}{n} \right) \right|=1-\prod_{h=0}^{k-1}\left(1-\frac{h}{n} \right)$$ for $n \in \mathbb{N}^*, k \leqslant n$ and since $1-\frac{h}{n} \in [0,1]$ for any $h$ satisfying $0 \leqslant h \leqslant k-1$ an application of the mentioned inequality yields $$1-\prod_{h=0}^{k-1}\left(1-\frac{h}{n} \right) \leqslant \sum_{h=0}^{k-1} \frac{h}{n}=\frac{k(k-1)}{2n}$$