I've been trying to proof this theorem. I think I can prove it by using mathematical induction but I'm not sure about my attempt. Any hints on how I can tackle this problem will be greatly appreciated.
Let $f_{1},f_{2},\cdot\cdot\cdot f_{n}$ be paths on $X$ such that $f_{i}(1)=f_{i+1}(0)$, for $i=1, 2, \cdot\cdot\cdot, n-1$. Then
$[f_{1}]\ast[f_{2}]\ast \cdot\cdot\cdot\ast[f_{n}]=[(((f_{1}\ast f_{2})\ast f_{3})\ast\cdot\cdot\cdot)\ast f_{n}].$
Proof. This proof is by induction on $n$.
If $n=2$, the problem comes down to proving that $[f_{1}]\ast[f_{2}]=[f_{1}\ast f_{2}]$ which follows inmediatly from the definition of $\ast$.
Suppose that the result holds $n=k$, we have
\begin{align*} [f_{1}]\ast[f_{2}]\ast\cdot\cdot\cdot\ast[f_{k}]=[(((f_{1}\ast f_{2})\ast f_{3})\ast\cdot\cdot\cdot)\ast f_{k}]. \end{align*}
Now let's prove that the results holds for $n=k+1$.
\begin{align*} [f_{1}]\ast[f_{2}]\ast\cdot\cdot\cdot\ast[f_{k+1}]&=([f_{1}]\ast[f_{2}]\ast\cdot\cdot\cdot\ast[f_{k}])\ast[f_{k+1}] \\ &=[(((f_{1}\ast f_{2})\ast f_{3})\ast\cdot\cdot\cdot)\ast f_{k}]\ast[f_{k+1}] \\ &=[((((f_{1}\ast f_{2})\ast f_{3})\ast\cdot\cdot\cdot)\ast f_{k})\ast f_{k+1}]\\ &=[(((f_{1}\ast f_{2})\ast f_{3})\ast\cdot\cdot\cdot)\ast f_{k+1}]. \end{align*}