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For $\theta$ = 2

$\Gamma$($\theta) ^2$ = [(2-1)!]$^2$ = 1

or

$\Gamma$($\theta) ^2$ = (2$^2$-1)! = 3

Which one is correct?

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  • $\begingroup$ The square is outside the parentheses, it applies to the entire function, not the argument. $\endgroup$ – Michael Burr May 10 '18 at 20:09
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Since $\Gamma(n)=(n-1)!,$ and since $\Gamma(\theta)^2 = [\Gamma(\theta)]^2,$ then when $\theta=2,$ we have

$$\Gamma(2)^2 =[(2-1)!]^2=1.$$

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