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")?


"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."
In fact $(1 + x^2)(1 + x^4)(1 + x^8)(1 + x^{16}) \cdots = 1 + x^2 + x^4 + x^6 + x^8 + x^{10} + \ldots = \frac{1}{1 - x^2}$ -- the first equality is just the generating-functional version of binary expansion, see e.g. this recent question: Demonstrate recursively that $\prod_{k = 0}^\infty (1 + x^{2^k}) = \frac{1}{1-x}$.  (Or maybe you mean the fact that you've lost a minus sign?  That's because the leftover factor is $x^{2^n} - 1$, which is approaching $-1$ as $n \to \infty$.)
