I'm willing to prove that $\Gamma(e+1) = e!$ is irrational, in this task i found that the integral:
$\large \Gamma(e+1) = \int _0^{\infty }\:e^{-x}x^edx$
Can be written as: $\large \int _0^{\infty }\:\sum \limits_{n=0}^{\infty}\frac{(-1)^nx^n}{n!}\:x^edx$, since $\large e^{-x} = \sum \limits_{n=0}^{\infty}\frac{(-1)^nx^n}{n!}$, so my derivation goes on:
$\large \int _0^{\infty }\:\left(\frac{1}{0!}-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+...\right)x^edx$, wich implies:
$\large \int _0^{\infty }\:\left(\frac{x^e}{0!}-\frac{x^{e+1}}{1!}+\frac{x^{e+2}}{2!}-\frac{x^{e+3}}{3!}+...\right)dx$, now we solve this basic integral to get $e!$ in terms of the limit:
$$\large \lim_{t\to\infty} \large \int _0^{t}\:\left(\frac{x^e}{0!}-\frac{x^{e+1}}{1!}+\frac{x^{e+2}}{2!}-\frac{x^{e+3}}{3!}+...\right)dx$$
Conclusion:
$$\large \lim_{t\to\infty} \left(\frac{t^{e+1}}{\left(e+1\right)0!}-\frac{t^{e+2}}{\left(e+2\right)1!}+\frac{t^{e+3}}{\left(e+3\right)2!}-\frac{t^{e+4}}{\left(e+4\right)3!}+...\right) = \large e!$$
So, there's any mistake in this limit for of $e!$? If there's no mistake, then any factorial of a number $\sigma$ can be written in this very same form?