If: $$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx$$ $$z\Gamma(z)=\Gamma(z+1)$$ $$\Gamma(z)=\mathcal \{Me^{-x}\}(z)$$ $$\Gamma(z)=\lim_{x\to\infty}(-x^{z-1})-(-0^ze^{-0})+z\int_0^\infty x^{z-1}e^{-x}dx$$ $$z!=\prod_{n=1}^\infty\Biggl[{1\over{1+{z\over{n}}}}\biggl(1+{1\over n}\biggl)^z\Biggl]\cdot$$ Where $\mathfrak R(z)>0$ and
$$\Gamma(z+1)=z!$$ Where $\mathfrak R(z)>0$ Then is
$$z! = \int_0^\infty x+1^{z-1}e^{-x+1}dx?$$ This is what I did:
$$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx$$ $$\Gamma(z+1)=z!$$ Then +1 and -1 cancel. Is this right?
Link to gamma function: https://en.wikipedia.org/wiki/Gamma_function
You mean $\Gamma(\color{blue}{z})=\int_0^\infty x^{z-1}e^{-x}dx$; $x$ is integrated over, so doesn't appear on the left-hand side. So $z!=\Gamma(z+1)=\int_0^\infty x^ze^{-x}dx$.