Proof by induction: For all $m \in \Z_{> 0}$, let $P \left({m}\right)$ be the proposition: :$\Gamma \left({m + \dfrac 1 2}\right) = \dfrac {\left({2 m}\right)!} {2^{2 m} m!} \sqrt \pi$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \Gamma \left({1 + \frac 1 2}\right)       | r = \frac 1 2 \Gamma \left({\frac 1 2}\right)       | c = Gamma Difference Equation }} {{eqn | r = \frac {\sqrt \pi} 2       | c = Gamma Function of One Half }} {{eqn | r = \frac 2 4 \sqrt \pi       | c =  }} {{eqn | r = \frac {2!} {2^{2 \times 1} 1!} \sqrt \pi       | c = Definition of Factorial }} {{eqn | r = \frac {\left({2 m}\right)!} {2^{2 m} m!} \sqrt \pi       | c = where $m = 1$ }} {{end-eqn}} and so $P(1)$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\Gamma \left({k + \dfrac 1 2}\right) = \dfrac {\left({2 k}\right)!} {2^{2 k} k!} \sqrt \pi$ Then we need to show: :$\Gamma \left({k + 1 + \dfrac 1 2}\right) = \dfrac {\left({2 \left({k + 1}\right)}\right)!} {2^{2 \left({k + 1}\right)} \left({k + 1}\right)!} \sqrt \pi$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \Gamma \left({k + 1 + \frac 1 2}\right)       | r = \left({k + \frac 1 2}\right) \Gamma \left({k + \frac 1 2}\right)       | c = Gamma Difference Equation }} {{eqn | r = \left({k + \frac 1 2}\right) \dfrac {\left({2 k}\right)!} {2^{2 k} k!} \sqrt \pi       | c = Induction Hypothesis }} {{eqn | r = \frac {\left({2 k + 1}\right) \left({2 k}\right)!} {2 \times 2^{2 k} k!} \sqrt \pi       | c = simplifying }} {{eqn | r = \frac {\left({2 k}\right)! \left({2 k + 1}\right) \left({2 k + 2}\right)} {2 \left({2 k + 2}\right) 2^{2 k} k!} \sqrt \pi       | c = multiplying top and bottom by $2 k + 2$ }} {{eqn | r = \frac {\left({2 k + 2}\right)!} {2^{2 k + 1} \left({2 \left({k + 1}\right)}\right) k!} \sqrt \pi       | c =  }} {{eqn | r = \frac {\left({2 \left({k + 1}\right)}\right)!} {2^{2 \left({k + 1}\right)} \left({k + 1}\right)!} \sqrt \pi       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Finally: {{begin-eqn}} {{eqn | l = \frac {\left({2 m}\right)!} {2^{2 m} m!}       | r = \frac {1 \times 2 \times 3 \times \cdots \times 2 m} {2^{2 m} \ 1 \times 2 \times 3 \times \cdots \times m}       | c =  }} {{eqn | r = \frac {1 \times 2 \times 3 \times \cdots \times \left({2 m - 1}\right) \times 2 m} {2^m \ 2^m \ \left({1 \times 2 \times 3 \times \cdots \times m}\right)}       | c =  }} {{eqn | r = \frac {1 \times 2 \times 3 \times \cdots \times \left({2 m - 1}\right) \times 2 m} {2^m \left({\left({2 \times 1}\right) \times \left({2 \times 2}\right) \times \left({2 \times 3}\right) \times \cdots \times 2 m}\right)}       | c =  }} {{eqn | r = \frac {1 \times 2 \times 3 \times \cdots \times \left({2 m - 1}\right) \times 2 m} {2^m \left({2 \times 4 \times 6 \times \cdots \times 2 m}\right)}       | c =  }} {{eqn | r = \frac {1 \times 3 \times 5 \times \cdots \times \left({2 m - 1}\right)} {2^m}       | c =  }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | ll= \forall m \in \Z_{>0}:       | l = \Gamma \left({m + \frac 1 2}\right)       | r = \frac {\left({2 m}\right)!} {2^{2 m} m!} \sqrt \pi       | c =  }} {{eqn | r = \frac {1 \times 3 \times 5 \times \cdots \times \left({2 m - 1}\right)} {2^m} \sqrt \pi       | c =  }} {{end-eqn}} {{qed}}