Creating a degree $n$ Taylor polynomial for $\sqrt{1+x}$ I have been asked to produce a general formula for the degree $n$ Taylor polynomial for $\sqrt{1+x}$ using a=0 as the point of approximation.
Given that $p_n(x)=f(a)+(x-a)f^\prime(a)+\frac{(x-a)^2}{2!}f^{(2)}(a)+...+\frac{(x-a)^n}{n!}f^{(n)}(a)=\sum_{j=0}^{n}\frac{(x-a)^j}{j!}f^{(j)}(a)$
$f(x)=\sqrt{1+x}=(1+x)^{1/2} \\ 
f^\prime (x)=\left ( \frac{1}{2} \right )(1+x)^{-1/2} \\ 
f^{(2)}(x)=\left (  \frac{1}{2}\right )\left ( \frac{-1}{2} \right )(1+x)^{-3/2} \\
f^{(3)}(x)=\left (  \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )(1+x)^{-5/2} \\
f^{(4)}(x)=\left (  \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )\left ( \frac{-5}{2} \right )(1+x)^{-7/2} \\
f^{(j)}(x)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)(1+x)^{(2j-1)/2} \\
f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$
I'm thinking I have calculated the j-th derivative correctly, however I can't seem to figure out how to simplify it and the rest of the steps into a compact summation.
Added:  It's been pointed out to me that the $(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$ above should be $\prod_{i=0}^{j-1}2i-1$
Thus $f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )\prod_{i=0}^{j-1}2i-1$.
If anyone can give me a hint on how to simplify this further it would be appreciated.

Note that $$1\cdot 3\cdot 5 \cdots (2j - 3) = \dfrac{(2j-3)!}{2\cdot 4 \cdot 6 \cdots (2j -2)} = \dfrac{(2j-3)!}{2^{j-1}\cdot 1\cdot 2 \cdot 3 \cdots (j -1)} = 2^{1-j}\dfrac{(2j-3)!}{(j-1)!}.$$
