Integral Form of Gamma Function equivalent to Euler Form/Proof 2

Proof
First we present a lemma:

Recall the definition of the partial Gamma function:
 * $\ds \map {\Gamma_m} x := \frac {m^x m!} {x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + m} }$

We have that:

We have that for large $t$:
 * $t^{x - 1} < t^{t / 2}$

and so as $m \to \infty$:
 * $\ds \int_m^\infty e^{-t} t^{x - 1} \rd t \to 0$

Then:

Now we have that as $m \to \infty$:
 * $\ds \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t \to 0$

so:
 * $\ds \int_0^\infty e^{-t} t^{x - 1} \rd t - \map {\Gamma_m} x = 0$

leading to:
 * $\ds \int_0^\infty e^{-t} t^{x - 1} \rd t = \lim_{m \mathop \to \infty} \dfrac {m^x m!} {x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + m} }$

as was to be demonstrated.