A variant to Newton's method

I must consider the iteration $x_{n+1}=x_n - \frac{f(x_n)}{f'(x_0)}$. I must find the constants $C$ and $s$ such that $e_{n+1}=Ce_n ^s$.
Where $e_n=x_n-r$ and $r$ is the root of f.

My attempt: $e_{n+1}=x_{n+1}-r=x_n-\frac{f(x_n)}{f'(x_0)}-r=e_n-\frac{f(x_n ) }{f' (x_0 )}$.

So I must rewrite $e_n-\frac{f(x_n ) }{f' (x_0 )}$ as $C e_n ^s$. Thus I think I must relate $e_n$ to $\frac{f(x_n)}{f(x_0)}$ but I don't know how to proceed.
Any idea?
Thank you.