I think I'm pretty close to getting this but can't quite figure out what do next!

From earlier in the question...

$\displaystyle \textrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} dt \sim 1 - \frac{e^{-x^2}}{\pi} \sum^{\infty}_{n=0} \Gamma(n + \frac{1}{2})\frac{(-1)^n}{x^{2n+1}}$ (*)

Show that erf(x) can be expressed in terms of the incomplete gamma function, and that the above asymptotic expansion agrees with the result obtained in Problem 6. [The formula $\displaystyle \Gamma(\frac{1}{2} + n)\Gamma(\frac{1}{2} - n) = (-1)^n \pi$ (**) for integer n may be useful.]

So I think I have to show the relation... $\displaystyle \gamma(\frac{1}{2}, x) = \sqrt{\pi} \textrm{erf}(\sqrt{\pi})$ (***) as that's the relation I've seen elsewhere.

The result obtained in question 6 was...

$\displaystyle \gamma(\alpha,x) \sim \Gamma(\alpha) - e^{-x}x^{\alpha - 1} \sum^{\infty}_{n=0}\frac{\Gamma(\alpha)}{\Gamma(\a lpha - n)}\frac{1}{x^n}$

And also...
$\displaystyle \gamma(\alpha,x) \sim \Gamma(\alpha) - e^{-x}x^{\alpha - 1} \sum^{\infty}_{n=0}\frac{(\alpha-1)(\alpha-2)\ldots(\alpha-n)}{x^n}$

Right so intro done!

Now what I've done...

From (*)
$\displaystyle \sqrt{\pi} \textrm{erf}(x) = 2 \int^x_0 e^{-t^2} dt \sim \sqrt{\pi} - \frac{e^{-x^2}}{\sqrt{\pi}} \sum^{\infty}_{n=0} \Gamma(n + \frac{1}{2})\frac{(-1)^n}{x^{2n+1}}$

$\displaystyle = \Gamma(\frac{1}{2}) - \frac{e^{-x^2}}{\sqrt{\pi}} \sum^{\infty}_{n=0} \Gamma(n + \frac{1}{2})\frac{(-1)^n}{x^{2n+1}}$
(This step seems to be showing that $\displaystyle \alpha = \frac{1}{2}$ which corresponds to (***))

Using (**)
$\displaystyle = \Gamma(\frac{1}{2}) - \sqrt{\pi} e^{-x^2} $$\displaystyle \sum^{\infty}_{n=0} \frac{1}{\Gamma(\frac{1}{2} - n)}\frac{1}{x^{2n+1}}$

$\displaystyle = \Gamma(\frac{1}{2}) - \sqrt{\pi} e^{-x^2} \sum^{\infty}_{n=0} \frac{x^{-(n+1)}}{\Gamma(\frac{1}{2} - n)}\frac{1}{x^n}$

But I'm stuck now. I don't even know if what I've done so far is right to be honest... Any help please!