I was reading about Gamma function, its properties, its applications, etc., and I found the following property:
$$\Gamma(x)=\lim_{n\to\infty}\frac{n!n^x}{(x)_{n+1}}.$$
I want to prove this property, so I used the definition of the Gamma function: $\Gamma(x)=\int_0^\infty{t^{x-1}e^{-t}dt}$; and the definition of $e^{-t}$: $e^{-t}=\lim_{n\to\infty}(1-\frac{t}{n})^n\, dt$. Substitute in $\Gamma(x)$, and simplify:
$$\Gamma(x)=\lim_{n\to\infty}\int_0^nt^{x-1}\left(1-\frac{t}{n}\right)^ndt.$$
Now, I am stuck here, and asking if there any way to continue in that way or not.
As @Gary commented, you could use the beta function. Another solution using the gaussian hypergeometric function is $$I_n=\int t^{x-1}\left(1-\frac{t}{n}\right)^n\,dt=\frac{t^x }{x}\, _2F_1\left(-n,x;x+1;\frac{t}{n}\right)$$ $$J_n=\int_0^n t^{x-1}\left(1-\frac{t}{n}\right)^n\,dt=\frac{n^x }{x}\, _2F_1\left(-n,x;x+1;1\right)=\frac{n^x }{x}\,\frac{\Gamma (n+1) \Gamma (x+1)}{\Gamma (n+x+1)}$$ Rewrite $$J_n=\frac{\Gamma (x+1)}x \Bigg[n^x \frac{\Gamma (n+1)}{\Gamma (n+x+1) }\Bigg]=\Gamma (x)\Bigg[n^x \frac{\Gamma (n+1)}{\Gamma (n+x+1) }\Bigg]$$ Now, using series for large values of $n$ $$\log\Bigg[n^x \frac{\Gamma (n+1)}{\Gamma (n+x+1) }\Bigg]=-\frac{x (x+1)}{2 n}+O\left(\frac{1}{n^2}\right)$$ $$J_n=\Gamma (x)\Bigg[1-\frac{x (x+1)}{2 n}+O\left(\frac{1}{n^2}\right)\Bigg]$$