Gamma Function with Big O I want to prove the following:
$\Gamma (z+1)= \sqrt {2 \pi} {z^{z+ \frac{1}{2}}}\,\,\, e^{-z} \,\,[1+ O(\frac{1}{z})]$ and that 
$\Gamma (z+1)= \sqrt {2 \pi} {z^{z+ \frac{1}{2}}}\,\,\, e^{-z} \,\,[1+ \frac{1}{12z}+ O(\frac{1}{z^{2}})]$ where $z$ is real and greater than zero.
I know how to show that 
$\Gamma ({n})=n!\approx  \sqrt {2 \pi} {n^{n+ \frac{1}{2}}}\,\,\, e^{-n} $ 
so I write 
$\Gamma ({n})=n!=  \sqrt {2 \pi} {n^{n+ \frac{1}{2}}}\,\,\, e^{-n}\, \,(1+ \frac{a_{1}}{n}+\frac{a_{2}}{n^{2}}+...)$ 
and since the formula is valid for $n+1$ too I also write 
$\Gamma ({n+1})=(n+1)! =  \sqrt {2 \pi} \,{(n+1)^{n+1 + \frac{1}{2}}}\,\,\,\,\, e^{-(n+1)}\, \,\,\,(1+ \frac{a_{1}}{n+1}+\frac{a_{2}}{(n+1)^{2}}\,\,+...)$ 
and since 
$(n+1)!=(n+1)n!$ 
I got  
$(1+ \frac{a_{1}}{n}+\frac{a_{2}}{n^{2}}+...)=(1+\frac{1}{n})^{n+\frac{1}{2}}\,\,\,e^{-1}\,\,(1+\frac{a_{1}}{n+1}+\frac{a_{2}}{(n+1)^{2}}\,\,...)$ 
when I plugged the right hand side of the $n!$ into the last equation. But I don't know how to get the coefficients $a_{i}$'s from this point. I appreciate for any help. Thanks.

What you wrote is essentially Stirling's formula. The proof of it for the case of $\Gamma$-function is not entirely trivial(for example you can see from my question here), and I suspect not possible in a short answer at here, even though you can take various short cuts here and there in the proof. 
You can check Ahlfors, Complex Analysis, around page 200. I think there should be a chapter on this at Stein's book as well. 
