Prove that the result of any such nested composition is independent of the placement of the parentheses.
So this is what I have so far.
Proof by induction I want to show that for any such choice for FnFn-1.....F1 equals the multiplication of the functions. I read in the book that the basic step should be for n=5.
Where to go now? I do not know
On
If a binary operation has the property that keeps staying the same as parentheses are reorganized, it is simply called associative, and the basic step is rather for $n=3$: $$f\circ(g\circ h)=(f\circ g)\circ h \ .$$ Let's assume we have $f_n\circ\dots\circ f_1$ somehow parenthesized, e.g. $(f_5((f_4f_3)f_2))f_1$. Then, for example, apply the replacement $(uv)w \to u(vw)$ to all possible instantiation of $(uv)w$, so that finally we get $f_{n+1}(..(..(f_2f_1)).\!.\!)$, in the example above it goes like e.g.: $$(f_5((f_4f_3)f_2))f_1 = (f_5(f_4(f_3f_2))f_1 = f_5((f_4(f_3f_2))f_1) =\\ = f_5(f_4((f_3f_2)f_1)) = f_5(f_4(f_3(f_2f_1)). $$
As a matter of notation, let $O_k^m$ denote the composition $F_m\circ(F_{m-1}\circ(\ldots\circ F_k)\cdots ))$ and let $X_k^m$ denote any compostion of the same $F$'s in the same order, but with some arbitrary parenthesisation.
We need to show $X_k^m=O_k^m$ for all $k\ge m\ge 1$. Use induktion on $n=k-m$, the cases $n\le 2$ being trivial (no parentheses).
Given some $X_k^m$ with $k-m>2$, it is of the form $X_k^m=X_{r+1}^m\circ X_k^r$ for some intermediate $r$. By induction hypotheses, $X_{r+1}^m=O_{r+1}^m$ and $X_k^r=O_k^r$. If $r+1=m$, we simply have $$X_k^m=X_m^m\circ X_k^{m-1}=O_{m}^m\circ O_k^{m-1}=F_m\circ O_k^{m-1}=O_k^m$$ and otherwise we have $$X_k^m=X_{r+1}^m\circ X_k^{r}=O_{r+1}^m\circ O_k^{r}=(F_m\circ O_{r+1}^{m-1})\circ O_k^{r}=F_m\circ (O_{r+1}^{m-1}\circ O_k^{r})=F_m\circ O_k^{m-1}=O_k^m.$$