why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$ 
Why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1)  \cdot  (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$?

I think the equality holds as formal power series. Expanding the LHS, one should get $(1+x+x^2 +\cdots )(1+x^2 + x^4+\cdots).$ But doing cancellation for the RHS gives that it equals $(x^2+1)/(x-1) \cdot (x^4 + 1)(x^8 + 1)\cdots,$ which clearly contradicts the equality. If however, one multiplies both sides by $x^2 - 1$, one gets an obvious equality after cancellation.

I was wondering if someone could explain what I'm doing wrong here (i.e. why I get two different "answers")?


It is a telescoping product.  Everything on the RHS cancels in pairs, except the $1-x$ and $1-x^2$, which appear in the denominator but not the numerator.
