For $j \in \{0,...,n-1\}$ is $(n-j)!(j+1)! \leq n!$ true?

Question

For $j \in \{0,...,n-1\}$ is $(n-j)!(j+1)! \leq n!$ true?

I mean $\dfrac{n!}{(n-j)!(j+1)!}$ doesn't have to be an integer. I need this inequality in another exercise, so Is it provable?

Answer 1

Hint: Compare $$ (n-j)!(j+1)!=1\cdot \ldots\cdot (n-j)\cdot 2 \cdot \ldots \cdot (j+1) $$ with $$ n! = 1\cdot\ldots\cdot (n-j)\cdot (n-j+1)\cdot\ldots n. $$

Answer 2

The claim is that $$ {n \choose j} \geq (j + 1). $$ provided $j \neq n$. The left hand side is the number of ways you can choose $j$ boxes out of a room with $n$ boxes.

Enumerate the boxes. Pick the $j + 1$st box (makes sense because $j \neq n$), and all boxes less than it, now you have $j + 1$ boxes; choose one to exclude. That gives $j+1$ different ways to pick $j$ boxes.