Proof

• By hypothesis, $x > 0$ is a positive real number.
• Assume $R > \epsilon > 0.$
• By definition of the gamma function, $\Gamma(x+1):=\int_0^\infty \exp(-t)t^{x}dt.$
• In particular, the improper integral of the Gamma function $G(x+1)$ is the limit $\epsilon\searrow 0,$ and $R\to\infty$ of the Riemann integral $\int_\epsilon^R \exp(-t)t^{x}dt.$
• By partial integration $$\int_\epsilon^R \exp(-t)t^{x}dt=\underbrace{-\exp(-t)t^{x}\;\Rule{1px}{4ex}{2ex}^{t=R}_{t=\epsilon}}_{=:A}+x\int_\epsilon^R \exp(-t)t^{x-1}dt.$$
• For $\epsilon\searrow 0,$ and $R\to\infty$, the term $A$ converges to $$\lim_{\substack{\epsilon\searrow 0,\\ R\to\infty}}=-\frac{R^x}{\exp( R)}+\frac{\epsilon^x}{\exp(\epsilon)}=-0+0=0,$$ and this yields the functional equation $$\Gamma(x+1)=x\Gamma(x).$$
• Because $$\Gamma(1)=\lim_{R\to\infty}\int_0^R\exp(-t)dt=\lim_{R\to\infty}(-\exp(-R)+1)=1,$$ for all natural numbers $n\in\mathbb N$ the above functional equation yields the interpolation of the factorial. $$\Gamma(n+1)=n\Gamma(n)=n(n-1)\Gamma(n-1)=n(n-1)\cdots 1\cdot\Gamma(1)=n!.$$

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983