Taylor-Series at point x = 0 We have
$$ f(x) = (1+x)^{1/3} $$ 
I have to find the taylor series of f at the point x =0. The problem I am facing is that I do not know how often do I have to derive f
$$ f'(x) = \frac{1}{3}(1+x)^{-2/3} $$
$$ f''(x) = -\frac{2}{9}(1+x)^{-5/3} $$
$$ f'''(x) = \frac{10}{27}(1+x)^{-8/3} $$
$$ f^{4}(x) = -\frac{80}{81}(1+x)^{-11/3}$$
As you can see, I can derive infinity times.
As you also can see, there is a certain pattern when you oberseve the derivatives. 
The pattern I recognised is something like that:
$$ f^{(n)}(x) = (-1)^{n+1}\frac{..}{3^n}(1+x)^{-\frac{3n-1}{3}} $$
I don't know what to put in ".." in the first fraction.
I also don't know if I am completely wrong or right. 
Thank you in anticipation. 
 A: From 
\begin{align*}
f'(x) &= \frac{1}{3}(1+x)^{-\color{blue}{2}/3}\\
f''(x) &= -\frac{2}{9}(1+x)^{-\color{blue}{5}/3} \\
f'''(x) &= \frac{10}{27}(1+x)^{-\color{blue}{8}/3} \\
f^{4}(x) &= -\frac{80}{81}(1+x)^{-11/3}\\
\end{align*}
we conclude the numerator is
\begin{align*}
\color{blue}{2}\cdot \color{blue}{5}\cdot \color{blue}{8}=\prod_{j=1}^3 (3j-1)=8!!!
\end{align*}
with $n!!!=n\cdot  (n-3)\cdot (n-6)\cdots $ the triple factorial.

In general $n$ we obtain
  \begin{align*}
 f^{(n)}(x) &= (-1)^{n+1}\frac{\prod_{j=1}^n (3j-1)}{3^n} (1+x)^{-\frac{3n-1}{3}} \\
 &= (-1)^{n+1}\frac{(3n-1)!!!}{3^n} (1+x)^{-\frac{3n-1}{3}} 
\end{align*}

A: Look at the following table, where the first column is the order of the derivative, and the second column the numerator of the fraction:
$$\begin{align}
1&\quad-1\\
2&\quad1\cdot2\\
3&\quad1\cdot2\cdot5\\
4&\quad1\cdot2\cdot5\cdot8\\
5&\quad1\cdot2\cdot5\cdot8\cdot11
\end{align}$$
Can you see the pattern?
A: $\prod\limits_{i=0}^{i=n-2}(2+3i) = 3^{n-1}\prod\limits_{i=0}^{i=n-2}(i+\frac{2}{3}) = 3^{n-1}(n-2+\frac{2}{3})!=3^{n-1}(n-\frac{4}{3})!$
