I need to prove that if $-1< x leq 0$, then the Taylor’s rest for $log(x+1)$ satisfies $$ |R_{n,0}| leq frac{|x|^{n+1}}{(x+1)(n+1)} $$

I have the next formula for the Taylor’s rest $R_{n,a}(x) = frac{f^{(n+1)}(s)}{n!}(x-s)^{n+1} $ for some $s in (a,x)$.

First thing I know, is that the n derivative of $log(x+1)$ looks like $$ f^{n}(x) = frac{(-1)^{n-1} (n-1)!}{(x+1)^{n}}$$ so the n+1 looks like $$f^{n+1}(x) = frac{(-1)^n n!}{(x+1)^{n+1}}$$

Then, the Taylor’s rest is $$ R_{n,0}(x) = frac{frac{(-1)^n n!}{(s+1)^{n+1}}}{n!}(x-s)^{n+1} = frac{(-1)^n}{(s+1)^n}(x-s)^{n+1}$$

Is it correct? And if it is, could you give me a hint to continue with the proof?