3 Oops, messed up Stirling's formula itself

$\pi$ shows up in at least two different ways related to factorials.

The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n /\sqrt{2 \times\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.

Also, $(-1/2)! = \Gamma(1/2) = \sqrt \pi.$ That can be seen from Euler's reflection formula,

$$\Gamma(x)\Gamma(1-x) = \frac {\pi}{\sin \pi x}$$

$\pi$ shows up in at least two different ways related to factorials.

The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n /\sqrt{2 \pi n}$ comes down to Wallis's Formula

$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$

which follows from the infinite product for sine

$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$

evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.

Also, $(-1/2)! = \Gamma(1/2) = \sqrt \pi.$ That can be seen from Euler's reflection formula,

$$\Gamma(x)\Gamma(1-x) = \frac {\pi}{\sin \pi x}$$

The $\pi$ in Stirling's Formula $n! \approx (\frac{n}{e})^n /\sqrt{2 \pi n}$ comes down to Wallis's Formula
$$\frac {\pi}{2} = \prod_{n=1}^\infty \frac{2n\times 2n}{2n-1 \times 2n+1}$$
$$\sin x = x \prod_{n=1}^\infty \bigg( 1 - \frac {x^2}{\pi^2 n^2}\bigg)$$
evaluated at $x=\frac\pi2$. I guess that comes down to the circle after all.