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For $z$ away from the poles, the gamma function satisfiesIts finite product is $m!(m+1{)}^z/[z(z+1)\cdots (z+m)]$. Splitting even and odd factors gives the gamma duplication formula.

The given Euler product for the gamma function has the finite approximantsaway from the poles. Split the denominator of ${\Gamma }_{2m+1}(2z)$ into even and odd factors. The resulting exact identity isThe right side does not depend on $z$, so its limit is a constant $C$. Evaluate the ratio at $z=1/2$: $\Gamma (1)=1$, and the gamma reflection formula, together with positivity on the positive real axis, gives $\Gamma (1/2)=\sqrt{\pi }$. HenceThis is the gamma duplication formula. The argument first applies where the factors are finite; the identity then holds as an identity of meromorphic functions, with values at poles interpreted accordingly.

In the integral defining the Beta function, put $t=(1+x)/2$ and then $u=x^2$. Symmetry givesThis proves the identity for $\mathrm{Re}z>0$. Both sides extend by analytic continuation to the same meromorphic function, equivalently by the beta--gamma identity and the Gamma duplication formula. The identity holds meromorphically for every $z\in \mathbb{C}$; both sides are finite for $z\notin \{0,-1,-2,\dots \}$ and have matching poles at the excluded points.