I really don't know where to start here. I have looked at "Rational Difference Equation" on Wikipedia, but found nothing useful.
The recurrence I am trying to solve is $$a_{n+1} = P/(P-1), \text{ where } P_{n-1} := \prod_{i=1}^{n}a_{i},$$ where $a_{1} \neq 1$.
The motivation for this comes from finding solutions in real numbers to $$\sum_{i=1}^{n} \alpha_{i} = \prod_{i=1}^{n} \alpha_{i},$$ where $n \in \mathbb{N}$. Of course $(\alpha, \frac{\alpha}{\alpha - 1})$ is a solution when $n = 2$ ($\alpha \neq 1$). So in general, we can find solutions for any $n$ by choosing $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n-1}$ to satisfy the sum/product relation above, and then letting $$\alpha_{n} = \frac{\alpha_{1}\alpha_{2}\cdots\alpha_{n-1}}{\alpha_{1}\alpha_{2}\cdots\alpha_{n-1} - 1}.$$ Finding explicitly the $\alpha_{i}$ in terms of $\alpha_{1} =: x$ is equivalent to the recurrence relation above (right?).
Edit. As lulu stated, letting $b_{n} := 1/a_{n}$ and $Q_{n} := \prod_{i=1}^{n} b_{i}$, we have $b_{n} = 1 - Q_{n-1}$. Noticing that $b_{n} = Q_{n}/Q_{n-1}$, we get the relation $$Q_{n} = Q_{n-1} - Q_{n-1}^{2} = Q_{n-1}(1 - Q_{n-1}),$$
which seems to be the logistic map. Perhaps there's another way to look at this problem?