For \(\Re s>0\) the integral \[ \Gamma(s)=\int_0^\infty e^{-t}t^{s-1}\dt \] converges absolutely, and it defines a holomorphic function on the right half-plane \(\ts{s\in\CC\st\Re s>0}\).
For \(t>0\), \(\abs{t^{s-1}}=\abs{e^{(s-1)\ln t}}=t^{\Re s-1}\). For \(0<\varepsilon<1\) put \[ \Gamma_\varepsilon(s)\coloneqq\int_\varepsilon^{1/\varepsilon}e^{-t}t^{s-1}\dt. \] The integrand is continuous in \((t,s)\) and holomorphic in \(s\), and the interval is compact, so each \(\Gamma_\varepsilon\) is entire.
Fix \(n\ge1\) and the strip \(A_n\coloneqq\ts{s\st \frac1n<\Re s<n}\). For \(s\in A_n\), \[ \abs{\Gamma(s)-\Gamma_\varepsilon(s)}\le\int_0^\varepsilon t^{1/n-1}\dt+\int_{1/\varepsilon}^\infty e^{-t}t^{n-1}\dt, \] using \(t^{\Re s-1}\le t^{1/n-1}\) for \(t\le1\) and \(t^{\Re s-1}\le t^{n-1}\) for \(t\ge1\). Both integrals are finite (so the integral defining \(\Gamma\) converges absolutely) and tend to \(0\) as \(\varepsilon\to0\), independently of \(s\). Hence \(\Gamma_\varepsilon\to\Gamma\) uniformly on each \(A_n\). A uniform limit of holomorphic functions on an open set is holomorphic, so \(\Gamma\) is holomorphic on each \(A_n\), and the strips \(A_n\) cover the right half-plane.