Rewriting product to a binomial I'm currently researching Wigner matrices. I wanted to calculate the moments of its spectral density. The probability density is 
$$\frac{1}{2\pi} \sqrt{4-x^2} \text{ for } x \in [-2,2] $$
I have found an expression for the $k^{th}$ moment by integrating
$$ \int_{-2}^2 \frac{x^k}{2\pi} \sqrt{4-x^2} \, dx.$$
This is $0$ if $k$ is uneven and 
$$ \frac{\prod _{i=1}^{\frac{k}{2}} \frac{4k - 8i +12}{k - 2i +4}}{k+1}$$ if $k$ is even. 
It is known that the moments are the Catalan numbers for k is even $$ \frac{1}{k+1}\binom{2k}{k}. $$ In Mathematica i found that my solutions are the Catalan numbers. But I can't figure out how to rewrite my product to the expression of the Catalan numbers. Which would be a way more intuitive expression. 
Are there any tricks one can use to write products to binomials?
 A: Since we have to consider $k$ even, we set $k=2l$ and show

The following is valid for $l\geq 1$:
\begin{align*}
\frac{1}{2l+1}\prod_{i=1}^l\frac{8l-8i+12}{2l-2i+4}=\frac{1}{l+1}\binom{2l}{l}
\end{align*}
  with $\frac{1}{l+1}\binom{2l}{l}$ the Catalan numbers.
We obtain
  \begin{align*}
\frac{1}{2l+1}\prod_{i=1}^l\frac{8l-8i+12}{2l-2i+4}
&=\frac{1}{2l+1}\prod_{i=0}^{l-1}\frac{8l-8i+4}{2l-2i+2}\tag{1}\\
&=\frac{1}{2l+1}\prod_{i=0}^{l-1}\frac{8i+12}{2i+4}\tag{2}\\
&=\frac{1}{2l+1}\prod_{i=0}^{l-1}\frac{2(2i+3)}{i+2}\\
&=\frac{2^l}{2l+1}\cdot\frac{(2l+1)!!}{(l+1)!}\tag{3}\\
&=\frac{2^l}{l+1}\cdot\frac{(2l)!}{l!\cdot(2l)!!}\tag{4}\\
&=\frac{1}{l+1}\cdot\frac{(2l)!}{l!l!}\tag{5}\\
&=\frac{1}{l+1}\binom{2l}{l}\\
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we shift the index $i$ by one

*In (2) we reverse the order of multiplication $i \rightarrow l-1-i$

*In (3) we use factorials and double factorials instead of the product symbol
\begin{align*}
(2l+1)!!=(2l+1)(2l-1)\cdots 5\cdot 3\cdot 1
\end{align*}

*In (4) we use the identity 
$
(2l)!=(2l)!!(2l-1)!!
$

*In (5) we use the identity
\begin{align*}
(2l)!!=(2l)(2l-2)\cdots 4\cdot 2=2^l\cdot l!
\end{align*}
Note: In OPs question the case $k$ even should be written as
\begin{align*}
\frac{1}{\frac{k}{2}+1}\binom{k}{\frac{k}{2}}
\end{align*}
in order to be consistent with the stated product expression.
A: Hint: First work on proving the following identity.
$$2^k \prod_{j=1}^k\,(2j-1) = \prod_{j=1}^k\,(k+j)$$
Note that the right-hand side is the numerator of $2k \choose k$.
