=== Lemma === First we establish: {{:Partial Gamma Function expressed as Integral/Lemma}} The proof continues by induction on $m$. For all $m \in \Z_{\ge 1}$, let $\map P m$ be the proposition: :$\displaystyle \map {\Gamma_m} x = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t       | r = \int_0^1 \paren {1 - t} t^{x - 1} \rd t       | c =  }} {{eqn | r = \int_0^1 t^{x - 1} \rd t - \int_0^1 t^x \rd t       | c =  }} {{eqn | r = \sqbrk {\frac {t^x} x}_0^1 - \sqbrk {\frac {t^{x + 1} } {x + 1} }_0^1       | c =  }} {{eqn | r = \frac 1 x - \frac 1 {x + 1}       | c =  }} {{eqn | r = \frac {x + 1 - x} {x \paren {x + 1} }       | c =  }} {{eqn | r = \frac 1 {x \paren {x + 1} }       | c =  }} {{eqn | r = \frac {1^x 1!} {x \paren {x + 1} }       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \map {\Gamma_k} x = k^x \int_0^1 \paren {1 - t}^k t^{x - 1} \rd t$ from which it is to be shown that: :$\displaystyle \map {\Gamma_{k + 1} } x = \paren {k + 1}^x \int_0^1 \paren {1 - t}^{k + 1} t^{x - 1} \rd t$ === Induction Step === This is the induction step: With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \paren {1 - t}^{k + 1}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\rd u} {\rd t}       | r = -\paren {k + 1} \paren {1 - t}^k       | c = Power Rule for Derivatives }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\rd v} {\rd t}       | r = t^{x - 1}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {t^x} x       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | r = \paren {k + 1}^x \int_0^1 \paren {1 - t}^{k + 1} t^{x - 1} \rd t       | o =        | c =  }} {{eqn | r = \paren {k + 1}^x \sqbrk {-\paren {k + 1} \paren {1 - t}^k \frac {t^x} x}_0^1 - \paren {k + 1}^x \int_0^1 \paren {-\paren {k + 1} \frac {t^x} x} \paren {\paren {1 - t}^k} \rd t       | c = Integration by Parts }} {{eqn | r = \frac {\paren {k + 1}^{x + 1} } x \int_0^1 \paren {1 - t}^k t^x \rd t       | c = $u v$ term vanishes }} {{eqn | r = \frac {\paren {k + 1}^{x + 1} } x \frac {\map {\Gamma_k} {x + 1} } {k^{x + 1} }       | c = Induction Hypothesis }} {{eqn | r = \frac {\paren {k + 1}^{x + 1} } x \frac {k^{x + 1} k!} {k^{x + 1} \paren {x + 1} \paren {x + 2} \cdots \paren {x + k} \paren {x + 1 + k} }       | c =  }} {{eqn | r = \frac {\paren {k + 1}^{x + 1} k!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + k} \paren {x + k + 1} }       | c =  }} {{eqn | r = \frac {\paren {k + 1}^x \paren {k + 1}!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + k} \paren {x + k + 1} }       | c =  }} {{eqn | r = \map {\Gamma_{k + 1} } x       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m \in \Z_{\ge 1}: \map {\Gamma_m} x = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t$ for $x > 0$. {{qed}}