If $p = n(n^2-1)(n^2-4)\cdots(n^2-100)$ then how is $p$ divisible by $(21!)$?

Question

I found this question in a book, I have no idea how to approach this problem. I can expand and write $$p = (n-10)(n-9)....n(n+1)....(n+10)$$ but I don't know what to do next.

Moreover the book says that since $p$ is divisible by $21!$, it will hence be divisible by $2!3!4!5!6!$, by $(5!)^4$, by $(10!)^2$ and by $10!11!$. Why is this so? If it is divisible by $21!$ I understand why, for instance it is divisible by $10!$ or $11!$ alone, but I cannot understand why it is divisible by $10!11!$.

How do I begin this problem? How can I figure out how $p$ is divisible by $21!$ when I have no idea what $n$ is?

Answer 1

Part of the question has already been answered by @J.G. $$\frac{(n+10)!}{(n-11)!}=21!\binom{n+10}{21}$$

Note that

\begin{align} \frac{21!}{11!}&=21\cdot20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\\ &=3\cdot7\cdot\color{red}2\cdot \color{red}{10} \cdot19 \cdot \color{red}9 \cdot2\cdot17\cdot\color{red}8\cdot\underbrace{2\cdot3}_{\color{red}6}\cdot\color{red}5\cdot\color{red}7\cdot2\cdot13\cdot\color{red}4\cdot\color{red}3\cdot\color{red}1\\ &=10!\times\ldots \end{align}

Answer 2

Your product is $\frac{(n+10)!}{(n-11)!}=21!\binom{n+10}{21}$.