ln(1+x) maclaurin series I found the first four derivatives of 
$ f(x) = ln(1+x) $
Then for all n > 1, $$ f^n(x) = \frac{(-1^{n+1})(n-1)!}{(1+x)^n } $$
So, $$ f^n(0) = (-1^{n+1})(n-1)! $$
By definition Maclaurin Series are defined as:
$$\sum_{n=0}^{\infty} \frac{f^n(0)}{n!}x^n$$
*Since the $ f^n(0) $ is only true when n > 1 then,
$$\sum_{n=1}^{\infty} \frac{f^{n+1}(0)}{n+1!}x^{n+1}$$
I continue by replacing each term and I get:
$$ ln(1+x) = - \sum_{n=1}^{\infty} \frac{(-1)^n}{n*(n+1)}x^{n+1}$$
I know that the answer should be :
$$ ln(1+x) = - \sum_{n=1}^{\infty} \frac{(-1)^n}{n}x^{n}$$
Where did I go wrong? 
"*" : Unsure about the step.
I'm really bad at taylor series so if you have any good sites for resources, it would be very much appreciated.
Thank you for any help and answers.

If $f(x)=\log(1+x)$, then we have
$$\begin{align}
f^{(1)}(x)&=(1+x)^{-1}\\\\
f^{(2)}(x)&=-(1)(1+x)^{-2}\\\\
f^{(3)}(x)&=(-1)(-2)(1+x)^{-3}\\\\
\vdots
f^{(n)}(x)&=(-1)(-2)\cdots (-(n-1))(1+x)^{-n}\\\\
\end{align}$$
Therefore, we can write
$$\begin{align}
f(x)&=\log(1+x)\\\\
&=\sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!}\\\\
&=\sum_{n=1}^\infty (-1)^{n-1} (n-1)!\frac{x^n}{n!}\\\\
&=\sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}
\end{align}$$
