I need to show, that $$ \Gamma\left(z\right) = \int_{0}^{\infty}t^{z - 1}\left[{\rm e}^{-t} -\sum_{\ell = 0}^{k - 1} \left(-1\right)^{\ell}\, \frac{t^{\ell}}{\ell!}\right]{\rm d}t $$ is for $-k < \operatorname{Re}\left(z\right)< -k + 1$.
And I don't find the right path to get it. Could someone give me a hint or help me $?$.
For $\Re(z) > 0$ $$\Gamma(z)-\sum_{k=0}^{K-1} \frac{(-1)^k}{k! (z+k)}=\int_0^\infty t^{z-1}(e^{-t}-1_{t< 1}\sum_{k=0}^{K-1} \frac{(-1)^k}{k!} t^k)dt\tag{1}$$ The RHS converges for $\Re(z) > -K$ and it gives the analytic continuation to this half-plane.
For $\Re(z)\in (-K,1-K)$, $$\sum_{k=0}^{K-1} \frac{(-1)^k}{k! (z+k)}= -\int_1^\infty t^{z-1}\sum_{k=0}^{K-1} \frac{(-1)^k}{k!} t^k dt\tag{2}$$ Adding $(1)$ and $(2)$ you get your result.