Prove Combinatoric Inequality

Question

Prove that $$\binom{n}{k}\ge\frac{n^k}{k!}\left( 1- \frac{k(k-1)}{n}\right)$$

for all $n\in\mathbb{N}$ and $k\in\mathbb{N_0}$ where $n\ge k.$

I think that I've tried rewriting this about a million different ways now and I'm still not getting anywhere. I've rewritten the inequality in a way that seems like it would be helpful, but I'm not sure how to proceed from here.

$$\frac{n!}{k!(n-k)!}\ge\frac{n^k}{k!} - \frac{n^{k-1}}{(k-2)!}$$

Answer 1

This is how I would begin :

$$\binom{n}{k}=\frac{1}{k!}n(n-1)\cdots (n-k+1)=\frac{n^k}{k!}\underbrace{\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)}_{A_{j,n}} $$

Now the name of the game is to show that $A_{j,n}$ is bigger than $\left(1-\frac{k(k-1)}{n}\right)$. Hint : develop the product.