A well presented solution from Richard of The Royal Hospital School reflected those of a number of other solvers including Kevin of Langley Grammar, Jeff from New Zealand and Andrei of Tudor Vianu School. Well done to all of you.

We are given that:

\begin{eqnarray} F_0(x) &=& 1/(1 - x) \\ F_n(x)
&=& F_0 (F_{n-1}(x)) \end{eqnarray}

This implies that: $$F_n(x) = 1 / (1 - F_{n-1}(x))$$ and can
be extended to: $$F_n(x) = 1 / (1 - (1 / (1 - F_{n-2}(x))))$$

Creating functions of $x$ when $n = 1, 2$ and $3$ gives:

\begin{eqnarray} F_1(x) &=&1 / (1 - F_0(x))\\
&=& 1 / (1 - (1 / (1 - x)))\\ &=& 1 / ((1 - x - 1)
/ (1 - x))\\ &=& 1(1 - x) / -x\\ &=& (1 - x) / -x\\
F_2(x) &=& 1 / (1 - F_1(x))\\ &=& 1 / (1 - ((1 - x)
/ -x))\\ &=& 1 / ((-x - 1 + x) / -x)\\ &=& -x /
-1\\ &=& x\\ F_3(x) &=& 1 / (1 - F_2(x))\\
&=& 1 / (1 - x)\\ &=& F_0(x) \end{eqnarray}
Therefore the function of x will repeat itself every three times.
\begin{eqnarray} F_0(x) &=& F_3(x)\\ F_1(x) &=&
F_4(x) \end{eqnarray} Etc. etc. So, to find $F_{2000}(x)$, we must
find the remainder given when 2000 is divided by three.$$Mod_3 2000
= 2$$ Thus, $F_{2000}(x) = F_2(x)$, and $F_2(x) = x$.

Therefore: $$F_{2}2000 = 2000$$