Factorials !!!

Using the facts that $n! = n(n-1)!$ and $(\frac{-1}{2})! = \sqrt{\pi},$ $(\frac{3}{2})! = \frac{3}{2}\times(\frac{1}{2})! \\ = \frac{3}{4}\times (\frac{-1}{2})! \\ = \boxed{\frac{3\sqrt{\pi}}{4}}$ Calculus is required to prove any of these values, but briefly, the reason that $(\frac{-1}{2})! = \sqrt{\pi}$ is that $\Gamma(n) = \int_0^\infty e^{-x} x^{n-1} dx \\ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$ Before you ask, the lower limit of integration changed to negative infinity since it is an even function; after substitution, the integral contains a $2$ which has been factored out.