proving algebraic expression involving rational function

If $$^nJ_r = \frac{(1-x^n)(1-x^{n-1})(1-x^{n-2})\cdots (1-x^{n-r+1})}{(1-x)(1-x^2)(1-x^3)\cdots (1-x^r)}$$

then prove that $$\displaystyle ^nJ_{n-r}=^nJ_{r}$$

what i try

$$^nJ_{n-r} = \frac{(1-x^n)(1-x^{n-1})(1-x^{n-2})\cdots (1-x^{r+1})}{(1-x)(1-x^2)(1-x^3)\cdots (1-x^{n-r})}$$

dod not kow how do i solve it help me please

• Set them equal and clear denominators. You get a true statement. Now just do the steps in reverse. – saulspatz Feb 11 at 14:27
• why not try some concrete examples, like $n=10, r=3$, to see what's going on? – Matthew Towers Feb 11 at 14:29
• It might help to write out explicitly a (slightly) non-trivial example, say ${}^7 J_3$ and see where the cancellations come from. – NickD Feb 11 at 14:30
• did not understand fully please explain me – jacky Feb 11 at 14:48

Let $$\displaystyle f(r)=(1-x)(1-x^2)(1-x^3)\cdots\cdots (1-x^r).$$
Then $$\displaystyle ^nJ_{r}=\frac{(1-x^n)(1-x^{n-1})\cdots (1-x^{n-r+1})\cdots (1-x^{n-r})(1-x^{n-r-1})\cdots (1-x)}{(1-x)(1-x^2)\cdots (1-x^{n-r})}\cdots \times \frac{1}{(1-x)(1-x^2)\cdots (1-x^{n-r})}.$$
$$\displaystyle ^nJ_{r}=\frac{f(n)}{f(r)\cdot f(n-r)}.$$
$$^nJ_{r}=^nJ_{n-r}.$$