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