How does $\frac{(n+1)\cdot n!\cdot|x|^n\cdot|x|}{9\cdot18\cdot27\cdots9(n+1)}\cdot \frac{9\cdot18\cdot27\cdots9n}{n!\cdot|x|^n}$ become $\frac19|x|$?

Question

Here's the solution for a question from my textbook. How do they get from the top step, to the bottom step?

$$\begin{align} &=\frac{(n+1)\cdot n!\cdot|x|^n\cdot|x|}{9\cdot18\cdot27\cdot\,\cdots\,\cdot 9(n+1)}\cdot \frac{9\cdot18\cdot27\cdot\,\cdots\,\cdot9n}{n!\cdot|x|^n} &\begin{array}{c} \text{Write $|x|^{n+1}$ as $|x|^n\cdot|x|$}\\ \text{and $(n+1)!$ as $(n+1)\cdot n!$.} \end{array} \\ &=\frac19|x| &\text{Simplify.} \end{align}$$

I'm aware it involves a lot of cancellation, but I'm unsure how the series's cancel out?

I've ran through it about 20 times, each trying a different way of cancelling out the series's but I can't get it to work.

Any help would be appreciated.

Answer 1

$$\frac{(n+1)\cdot n! \cdot |x|^n \cdot|x|}{(9\cdot 18\cdot \ldots \cdot 9n) \cdot9(n+1)}\cdot \frac{9\cdot 18\cdot \ldots \cdot 9n}{n! \cdot |x|^n}=\frac{(n+1)\cdot n! \cdot |x|^n \cdot|x|}{9\cdot (n+1)}\cdot \frac{1}{n!\cdot |x|^n}\\ =\frac{(n+1)\cdot |x|}{9(n+1)}=\frac{|x|}{9}$$