Getting the volume of a sphere from an infinite product I read an article about squeezing pi from a menger sponge. You can see the original here: Squeezing pi from a menger sponge
The author finds an infinite product which, he says, approximates the volume of a sphere with radius 1. The author mentions that the infinite product he found can be written in terms of the Wallis product. The equation given is:
$$8\prod_{n=1}^\infty \frac{(2n+1)^3-3(2n+1)+2}{(2n+1)^3}=\frac{4\pi}{3}$$ 
I do not know how to get the left hand side of the above equation in terms of the Wallis product. Hence, I do not know how to verify the equation. Please help.
For reference, the Wallis product is given by:
$$\prod_{n=1}^\infty \frac{2n}{2n-1}\cdot \frac{2n}{2n+1}=\frac{\pi}{2}$$ 
Suggestions?

You can derive this result as follows:
First, notice that:
$$(2n+1)^3-3(2n+1)+2 = 12n^2 + 8n^3 = 4n^2(2n+3)$$
Then, the product in your expression can be written as:
$$
\begin{align*}
\prod_{n=1}^\infty \frac{(2n+1)^3-3(2n+1)+2}{(2n+1)^3} &= \prod_{n=1}^\infty \frac{4n^2(2n+3)}{(2n+1)(2n+1)(2n+1)} \overset{1}{=} \prod_{n=1}^\infty \frac{4n^2(2n+3)}{(2n-1)(2n+1)(2n+1)} \\ &\overset{2}{=} \frac{1}{3}\prod_{n=1}^\infty \frac{4n^2(2n+3)}{(2n-1)(2n+1)(2n+3)} = \frac{1}{3}\prod_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1} = \frac{\pi}{6}
\end{align*}
$$
In (1) I just included 2*1 - 1 = 1 in the product of the first term, effectively shifting the starting place once to the left. Similarly, in (2) I shifted the starting place of the product for the third term once to the right, putting the remaining 3 outside.
Multiplying this by 8 gives you the result you are looking for!
