Methods to solving $a_{n+1} = \frac{a_{n}}{a_{n} - 1}$

Question

I have looked on the Wiki page for recurrence relation, and there is a section titled "rational difference equations", which fits my problem of solving $$a_{n+1} = \frac{a_{n}}{a_{n}-1}.$$ But the method on that page does not seem to work. Can anyone point me in the right direction?

Edit: I just realized that this is not useful to what I was trying to do! However it is still interesting to me, and I got some great answers. Thank you. I have posted the real question I am trying to ask in a different thread.

Answer 1

Hint (assuming $a_0 \ne 0\,$): write it as $\,\dfrac{1}{a_{n+1}} = 1 - \dfrac{1}{a_n}\,$ then it becomes a linear recurrence in $\,\dfrac{1}{a_n}\,$.

Answer 2

neither fixpoint is stable. If you have exactly $a_0 = 0,$ it stays there forever. If $a_0 = 2,$ it stays there forever. Otherwise it wanders all over. Let me put in a picture and then look for 2-cycles and 3-cycles...

CORRECTION: it has index 2. You ALWAYS HAVE $$ a_{n+2} = a_n. $$ If $a_0 \neq 0,1,2,$ then your system bounces back and forth between a pair of values forever.