Mittag-Leffler's expansion theorem states that $$f(z)=f(0)+\sum_{n=1}^\infty \text{Res}_{z=z_k}\left\{\frac{1}{z-z_k}+\frac{1}{z_k}\right\}$$ For gamma function $\Gamma(z)$ $$\Gamma(z)+\frac{1}{z}=1+\sum_{n=1}^\infty \frac{(-1)^n}{n!}\left(\frac{1}{z+n}-\frac{1}{n}\right)=1+\sum_{n=0}^\infty\frac{(-1)^n}{n!(z+n)}-\sum_{n=1}^\infty \frac{(-1)^n}{n(n)!}$$
But this doesn't simplifies to what I have to prove. That is $$\Gamma(z)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(z+n)}+\int_1^\infty t^{z-1}e^{-t}dt$$ I get the first term right!, but I'm not sure how does this integral shows up here. Please help me with this
I think there's no point in trying to invoke the general Mittag-Loeffler result. Just break the Euler integral into $\int_0^1$ and $\int_1^\infty$. In the $\int_0^1$, expand $e^{-t}$ in the usual power series. Interchange integral and sum, and you get the result you're aiming for.