From the Gamma Function Extends Factorial: :$f \left({x + 1}\right) = \begin{cases} 1 & : x = 0 \\ x f \left({x}\right) & : x > 0 \end{cases}$ From Gamma Function is Continuous on Positive Reals, $\Gamma$ is positive and continuous on $\R_{>0}$. From Log of Gamma Function is Convex on Positive Reals, $\ln \mathop \circ \Gamma$ is convex on $\R_{>0}$. It remains to be shown that $\Gamma$ is the ''only'' real function which satisfies these conditions. Let $s, t \in \R_{>0}$ such that $s \le t \le s + 1$. Let $t = \alpha s + \beta \left({s + 1}\right)$ where $\alpha, \beta \in \R_{>0}, \alpha + \beta = 1$. Then: :$t = \left({\alpha + \beta}\right) s + \beta = s + \beta$ and so: :$\beta = t - s$ Thus: {{begin-eqn}} {{eqn | l = \ln \left({f \left({t}\right)}\right)       | o = \le       | r = \alpha \ln \left({f \left({s}\right)}\right) + \beta \ln \left({f \left({s + 1}\right)}\right)       | c = $\ln \mathop \circ f$ is convex on $\R_{>0}$ }} {{eqn | ll= \leadsto       | l = f \left({t}\right)       | o = \le       | r = \left({f \left({s}\right)}\right)^\alpha \left({f \left({s + 1}\right)}\right)^\beta       | c =  }} {{eqn | r = \left({f \left({s}\right)}\right)^\alpha \left({s f \left({s}\right)}\right)^\beta       | c =  }} {{eqn | r = s^\beta f \left({s}\right)       | c = as $\alpha + \beta = 1$ }} {{eqn | n = 1       | r = s^{t - s} f \left({s}\right)       | c = as $\beta = t - s$ }} {{end-eqn}} Because $s \le t \le s + 1$, it follows that $t - 1 \le s \le t$. Substituting as appropriate in $(1)$: {{begin-eqn}} {{eqn | l = f \left({s}\right)       | o = \le       | r = \left({t - 1}\right)^{s - t + 1} f \left({t - 1}\right)       | c =  }} {{eqn | n = 2       | r = \left({t - 1}\right)^{s - t} f \left({t}\right)       | c =  }} {{end-eqn}} Combining $(1)$ and $(2)$: {{begin-eqn}} {{eqn | l = \left({t - 1}\right)^{t - s} f \left({s}\right)       | o = \le       | r = f \left({t}\right)       | c =  }} {{eqn | n = 3       | o = \le       | r = s^{t - s} f \left({s}\right)       | c =  }} {{end-eqn}} Let $x \in \R$ such that $0 < x \le 1$. Let $n \in \N$. Let: :$s = n + 1$ :$t = x + n + 1$ Then substituting in $(3)$: {{begin-eqn}} {{eqn | l = \left({x + n}\right)^x f \left({n + 1}\right)       | o = \le       | r = f \left({x + n + 1}\right)       | c =  }} {{eqn | o = \le       | r = \left({n + 1}\right)^x f \left({n + 1}\right)       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \left({x + n}\right)^x n!       | o = \le       | r = \left({x + n}\right) \left({x + n - 1}\right) \cdots x f \left({x}\right)       | c =  }} {{eqn | o = \le       | r = \left({n + 1}\right)^x n!       | c =  }} {{eqn | ll= \leadsto       | l = \left({1 + \frac x n}\right)^x       | o = \le       | r = \frac {\left({x + n}\right) \left({x + n - 1}\right) \cdots x f \left({x}\right)} {n^x n!}       | c =  }} {{eqn | o = \le       | r = \left({1 + \frac 1 n}\right)^x       | c =  }} {{end-eqn}} Thus when $0 < x \le 1$: : $\displaystyle f \left({x}\right) = \lim_{n \mathop \to \infty} \frac {x \left({x + 1}\right) \cdots \left({x + n}\right)} {n^x n!}$ which is the Euler form of the Gamma function. The general case is deduced by the use of: :$f \left({x + 1}\right) = x f \left({x}\right)$ {{qed}}