Partial Gamma Function expressed as Integral

Theorem
Let $m \in \Z_{\ge 1}$.

Let $\Gamma_m \left({x}\right)$ denote the partial Gamma function, defined as:
 * $\displaystyle \Gamma_m \left({x}\right) := \frac {m^x m!} {x \left({x + 1}\right) \left({x + 2}\right) \cdots \left({x + m}\right)}$

Then:
 * $\displaystyle \Gamma_m \left({x}\right) = \int_0^m \left({1 - \frac t m}\right)^m t^{x - 1} \, \mathrm d t$

for $x > 0$.

Lemma
First we establish:

The proof continues by induction on $m$.

For all $m \in \Z_{\ge 1}$, let $P \left({m}\right)$ be the proposition:
 * $\displaystyle \Gamma_m \left({x}\right) = m^x \int_0^1 \left({1 - t}\right)^m t^{x - 1} \, \mathrm d t$

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \Gamma_k \left({x}\right) = k^x \int_0^1 \left({1 - t}\right)^k t^{x - 1} \, \mathrm d t$

from which it is to be shown that:
 * $\displaystyle \Gamma_{k + 1} \left({x}\right) = \left({k + 1}\right)^x \int_0^1 \left({1 - t}\right)^{k + 1} t^{x - 1} \, \mathrm d t$

Induction Step
This is the induction step:

With a view to expressing the primitive in the form:
 * $\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$

let:

and let:

Then:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall m \in \Z_{\ge 1}: \Gamma_m \left({x}\right) = \int_0^m \left({1 - \frac t m}\right)^m t^{x - 1} \, \mathrm d t$

for $x > 0$.