Gamma Function of One Half/Proof 4

Theorem

$\map \Gamma {\dfrac 1 2} = \sqrt \pi$


Proof


Due to the organization of pages at $\mathsf{Pr} \infty \mathsf{fWiki}$, this argument is circular.
In particular: This uses Gamma Function of Positive Half-Integer in its proof, which itself uses Gamma Function of One Half, so it's circular.
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\(\ds \map \Gamma {\frac 1 2}\) \(=\) \(\ds \frac {0!} {2^0 0!} \sqrt \pi\) Gamma Function of Positive Half-Integer
\(\ds \) \(=\) \(\ds \sqrt \pi\) Factorial of Zero

$\blacksquare$

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