The following proof is given in Introduction to Algebraic Topology by Rotman
Now I understand the proof except for the following line
Since $S^1$ is a multiplicative group there is a telescoping product in $S^1$: $$f(x) = f(x_0)[f(x_0)^{-1}f(x_1)][f(x_1)^{-1}f(x_2)] \dots [f(x_{n-1}^{-1}f(x_n)]$$
What I doesn't understand is how does $f(x) = f(x_0)[f(x_0)^{-1}f(x_1)][f(x_1)^{-1}f(x_2)] \dots [f(x_{n-1}^{-1}f(x_n)]$, since it seems that the right hand side (if we remove the brackets) is $ f(x_0)f(x_0)^{-1}f(x_1)f(x_1)^{-1}f(x_2) \dots f(x_{n-1}^{-1}f(x_n) = e$ where $e$ is the identity on the multiplicative group $S^1$ and $f(x)$ doesn't necessarily equal $e$.
The last term f(x$_n$) remains in the product since its inverse is not in the product and it says that x$_n$ = x.