For $\sigma \in \mathbb{R},$ define $\Omega_{\sigma} = \{ s\in \mathbb{C}\;|\; Re(s) > \sigma\}$. On $\Omega_0$, define $\Gamma(s)$ by \begin{align*} \Gamma(s) = \int_{0}^{\infty}x^{s-1}e^{-x}dx \end{align*} I want to show that $\Gamma(s+1) = s\Gamma(s)$. Below is my attempt, which utilizes integration by parts: \begin{align*} \Gamma(s+1) & = \int_{0}^{\infty}x^{s}e^{-x}dx \\ &= -x^se^{-x}+ \int_{0}^{\infty}sx^{s-1}e^{-x}dx \\ &= -x^se^{-x} + s \Gamma(s) \end{align*} I can't find a way to get rid of the extra $-x^se^{-x}$ term. Am I approaching this in the wrong way? I would appreciate any help on this.
Also how can we show that there exists a unique holomorphic function $F: \mathbb{C} \setminus \{-k\;|\; k \in \mathbb{N}\cup 0\} \to \mathbb{C}$ with $F=\Gamma$ on $\Omega_{0}$? The book I'm using says to use the above identity that $\Gamma(s+1) = s\Gamma(s)$ and iteratively extend to $\Omega_{-l}\setminus \{-k\;|\; k \in \mathbb{N}\cup 0\}$ for $l\in \mathbb{N}$.
You just forgot to evaluate the term $-xe^{-x}|_0^\infty = 0$