I want to prove, the Laurent expansion of gamma function. \begin{align} \Gamma(z) = \frac1z-\gamma+\frac12\left(\gamma^2+\frac {\pi^2}6\right)z-\frac16\left(\gamma^3+\frac {\gamma\pi^2}2+2 \zeta(3)\right)z^2+O(z^3). \end{align}
First, my guess of obtaing above expansion, is starting from the definitions of gamma function \begin{align} \Gamma(z) &= \int_0^{\infty} dt e^{-t} t^{z-1} \\ & = \int_1^\infty dt e^{-t}t^{z-1} + \int_0^1dt e^{-t} t^{z-1} \\ & = \int_1^\infty dte^{-t}t^{z-1} + \int_0^1 dt t^{z-1} \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}t^n \\ & = \int_1^\infty dt e^{-t}t^{z-1} + \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\frac{1}{z+n} \end{align} This only gives the gamma function as a function of $\frac{1}{z}$...
Or should I start with \begin{align} \Gamma(z) = \lim_{n \rightarrow \infty} \frac{n! n^z}{z(z+1) \cdots(z+n)} \end{align}
An idea for you to develop:
The Weierstrass Formula tells us that
$$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac zn\right)e^{-z/n}$$
Now take logarithms on both sides to get a more or less well known relation:
$$-\log\Gamma(z)=\log z+\gamma z+\sum_{n=1}^\infty\left[\log\left(1+\frac zn\right)-\frac zn\right]$$
Now differentiate the above to get the logarithmic derivative of the Gamma function:
$$\frac{\Gamma'(z)}{\Gamma(z)}=-\frac1z-\gamma-\sum_{n=1}^\infty\frac1n\left[\frac n{z+n}-1\right]=-\frac1z-\gamma+\sum_{n=1}^\infty\frac z{n(z+n)}$$
and etc. You can try to integrate the $\;-\dfrac1z\;$ term into the series, too.