Show that $(n-j)! \leq \frac{n!}{(j+1)!}$

Question

I have to show that $$(n-j)! \leq \frac{n!}{(j+1)!}$$

I tried to develop $(n-j!)$ as $\frac{n!}{n(n-1)\cdot\cdot\cdot(n-j+1)}$ but I have no clue after that because I end up with $$\frac{1}{n(n-1)\cdot\cdot\cdot(n-j+1)} \leq \frac{1}{(j+1)!}$$

$${n(n-1)\cdot\cdot\cdot(n-j+1)} \geq {(j+1)!}$$

Answer 1

You could also just induct on $n$. The base case for $n = j+1$ is easy. The induction step isn't difficult either, as $$ (n+1-j)! = (n+1-j) (n-j)! \leq (n+1-j) \frac{n!}{(j+1)!} \leq (n+1) \frac{n!}{(j+1)!} = \frac{(n+1)!}{(j+1)!}. $$

Answer 2

If $j\geq 0$ and $j<n$ then $n+1 \leq {n+1\choose j+1}$, thus

$$1\leq {1\over n+1}{n+1\choose j+1} = {n! \over (j+1)!(n-j)!}$$