Gamma function in the complex plane

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Prove the formula

Γ(α)=∑k=0∞(−1)kk!(k+α)+∫1+∞xα−1e−xdx,α>0 \Gamma (\alpha ) = \sum _{k = 0}^{\infty } \frac{(-1)^{k}}{k !(k+\alpha )}+\int _{1}^{+\infty } x^{\alpha -1} e^{-x} d x, \quad \alpha > 0
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The Γ\Gamma-function

Γ(s)=lim⁡n→∞nsn!s(s+1)⋯(s+n) \Gamma (s) = \lim _{n \to \infty } \frac{n^{s} n!}{s(s+1)\cdots (s+n)}

can be written as an integral (Problem 31). Use this fact to prove that Γ(s)\Gamma (s) does not have any zeroes.

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Remark.
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Γ(s+1)=sΓ(s)\Gamma (s+1) = s\Gamma (s) (Problem 142).

The entire function 1Γ(z)\dfrac {1}{\Gamma (z)} assumes each value infinitely often in the half-strip

−d<Im(z)<d,Re(z)<0(d arbitrary). -d<\mathrm{Im}\left(z\right)<d, \quad \mathrm{Re}\left(z\right)<0 \quad (d \text{ arbitrary}).
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Euler's gamma function Γ(z)\Gamma (z) is defined for real z>0z > 0 by the equality

Γ(z)=∫0∞tz−1e−tdt \Gamma (z) = \int _{0}^{\infty } t^{z-1} e^{-t} d t

Prove that the function Γ(z)\Gamma (z) can be regularly continued to the entire complex plane, with the exception of the points z=0,−1,−2,…z = 0,-1,-2, \ldots.

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References

[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Функции многих переменных. 2-е изд., ФИЗМАТЛИТ, 2003.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.