Question related to deriving a result based on Stirling s formula I have been given a problem which I am unable to derive. 

Problem-> Using Stirling approximation formula derive that $\Gamma(x +1/n)$  ~ $ x ^{1/n} \Gamma(x) $  when x -> $\infty$ , ~ means asmyptotic to and x belongs to complex numbers and n belongs to positive integers. 

Attempt-> I used the formula $\Gamma(z+1/n) = {(\frac {2π} {z} )}^{1/2} {(\frac{z} {e}) }^{n} ( 1 +O(1/z)) $ 
But when put x = x+1/n and tend x to infinity , ${(\frac {2π} {x} )}^{1/2}$  tends to 0 . Do, I am doing some mistake or some other concept need to be used. 
I admit that I feel uncomfortable using Stirling approximations. 
Can someone please tell the right way to derive it. 
 A: Stirling is the asymptotic formula
$$\Gamma(x)\sim\sqrt{\frac{2\pi}x}e^{-x}x^x\tag1.$$
Take some positive number $a$ (you want $a=1/n$) and substituting in $(1)$
gives
$$\Gamma(x+a)\sim\sqrt{\frac{2\pi}{x+a}}e^{-x-a}(x+a)^{x+a}\tag2.$$
Divide $(2)$ by $(1)$ to give
$$\frac{\Gamma(x+a)}{\Gamma(x)}\sim\sqrt{\frac{x}{x+a}}e^{-a}
\frac{(x+a)^{x+a}}{x^x}\tag3.$$
As
$$\lim_{x\to\infty}\sqrt{\frac{x}{x+a}}=1,$$
$(3)$ simplifies to
$$\frac{\Gamma(x+a)}{\Gamma(x)}\sim e^{-a}
\frac{(x+a)^{x+a}}{x^x}=e^{-a}(x+a)^a\left(\frac{x+a}{x}\right)^x\tag4.$$
Now $(x+a)^a\sim x^a$ and
$$\lim_{x\to\infty}\left(\frac{x+a}{x}\right)^x
=\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x=e^a$$
(a standard limit). Therefore $(4)$ simplifies to
$$\frac{\Gamma(x+a)}{\Gamma(x)}\sim x^a.$$
A: Just use the plain formula
$$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \log \left(\frac{2 \pi
   }{p}\right)+O\left(\frac{1}{p}\right)$$ Use it twice and continue with Taylor series
$$\log \left(\Gamma \left(x+\frac{1}{n}\right)\right)-\log (\Gamma (x))=\frac{\log(x)}n+O\left(\frac{1}{x}\right)$$ Exponentiate both sides to get
$$\frac{\Gamma \left(x+\frac{1}{n}\right) } {\Gamma (x) }\sim x^{1/n}$$
