Prove that $\lim_{\ r\ \to \ \infty} \dfrac{r! r^x}{x(x+1)(x+2) \dots (x+r)} = \int_{0}^{\infty} t^{x-1} e^{-t} dt $ From Havil & Dyson, "Gamma: Exploring Euler's Constant", section 6.1 I can't prove the following Euler's theorem : 

... on 13 October 1729, Euler had already proposed to Goldbach the 
  definition $$\Gamma (x) = \lim_{r\ \to \ \infty} \Gamma_r (x)
$$ where $$\Gamma_r (x) = \dfrac{r! r^x}{x(x+1)(x+2) \dots (x+r)}$$ [which is valid whenever $x$ is not zero or negative integer]. 

After 1 hours and 40 minutes of attempt I can't turn $\lim_{r\ \to \ \infty} \Gamma_r (x)$ to the following definition of $\Gamma (x)$ that I know, i.e. $$\Gamma (x) = \int_{0}^{\infty} t^{x-1} e^{-t} dt, \ \ \ x>0. $$ 
So why $$\lim_{r\ \to \ \infty} \dfrac{r^x}{x(1+ \frac{x}{1})(1+\frac{x}{2}) \dots (1+\frac{x}{r})} = \int_{0}^{\infty} t^{x-1} e^{-t} dt , \ \ \ x>0 \ \text{?} $$
EDIT : For the special case when $x$ is a positive integer we will have the problem of showing $$\lim_{r\ \to \ \infty} \dfrac{r!r^n}{(n+r)!} =1 $$ to be valid.  
 A: The gamma function is well-defined by
$$\Gamma(z) = \lim_{r \to \infty} \Gamma_r(z),$$
as it can be shown that the limit exists as a convergent infinite product when $z$ is not zero or a negative integer.
We then can derive the integral form of the gamma function by first relating Euler's definition to the beta function.
The beta integral is defined as
$$B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt.$$
This satisfies
$$B(x,y) - B(x+1,y)= B(x,y+1) = \int_0^1t^{x-1}(1-t)^y \, dt.$$
Integrating by parts, we get
$$B(x,y+1) = \frac{y}{x}B(x+1,y)$$
Hence,
$$B(x,y) = B(x,y+1) + B(x+1,y) =  B(x,y+1) + \frac{x}{y}B(x,y+1) \\ \implies B(x,y) = \frac{x+y}{y}B(x,y+1).$$
Now using this recurrence relation $r$ times we find
$$B(x,y) = \frac{(x+y)(x+y+1) \ldots (x + y + r)}{y(y+1) \ldots (y +r)}B(x,y+r+1) \\ =\frac{\Gamma_r(y)}{\Gamma_r(x+y)}r^{x}B(x,y+r+1).$$
Note that
$$r^{x}B(x,y+r+1) = r^{x}\int_0^1t^{x-1}(1-t)^{y+r} \, dt.$$
Changing variables with $s = rt$ we get
$$r^{x}B(x,y+r) = \int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{y+r} \, ds = \int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}\, ds. $$
Hence,
$$B(x,y) = \frac{\Gamma_r(y)}{\Gamma_r(x+y)}\int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}\, ds.$$
We can progress towards representing the gamma function as an integral over $[0, \infty),$ by taking the limit of both sides of this equation as $r \to \infty$. The LHS will remain unchanged since there is no explicit dependence on $r$.
First, note that
$$\int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}\, ds = \int_0^{\infty}s^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}1_{[0,r]}\, ds , $$
and by the Lebesgue dominated convergence theorem
$$\lim_{r \to \infty}\int_0^rs^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}\, ds \\ = \int_0^{\infty} \lim_{r \to \infty}s^{x-1}\left(1-\frac{s}{r}\right)^{r} \left(1-\frac{s}{r}\right)^{y}1_{[0,r]}\, ds  \\ = \int_0^\infty s^{x-1}e^{-s}\, ds.$$
Since $\lim_{r \to \infty} \Gamma_r(z) = \Gamma(z)$, we find
$$B(x,y) = \frac{\Gamma(y)}{\Gamma(x+y)}\int_0^\infty s^{x-1}e^{-s}\, ds.$$
Finally, with $y = 1$ and using $\Gamma(1) = 1$ and $\Gamma(x+1) = x\Gamma(x),$ we find
$$\begin{align} \int_0^\infty s^{x-1}e^{-s}\, ds &= \frac{\Gamma(x+1)}{\Gamma(1)}B(x,1) \\ &= \frac{x\Gamma(x)}{1}\int_0^1t^{x-1} \, dt \\ &= \,\,\,\Gamma(x) \end{align}$$
