An identity involving the Gamma function I am trying to prove the following identity. Let $k\in\mathbb{N}$, $k\geq 2$. Then
$$
\sum_{i=1}^{k-1}\binom{k}{i}\prod_{l=2}^{i}(4l-6)\prod_{l=2}^{k-i}(4l-6)=\prod_{l=2}^{k}(4l-6).
$$
There is numerical evidence that the equation holds, also I have checked the identity for small values of $k$. Dividing the equation by $k!\cdot4^{k-2}$ and using the definition of the Gamma function it is easy to see that it is equivalent to show that
$$
\sum_{i=1}^{k-1}\frac{\Gamma(i-1/2)}{\sqrt{\pi}\Gamma(i+1)}\frac{\Gamma(k-i-1/2)}{\sqrt{\pi}\Gamma(k-i+1)}=4\prod_{l=2}^{k}\frac{\Gamma(k-1/2)}{\sqrt{\pi}\Gamma(k+1)}.
$$
I have tried to show this equation by induction, but I got stuck. Also, observing that 
$$
\prod_{l=2}^{k}(4l-6)=\frac{(2(k-1))!}{(k-1)!}
$$
is the highest order term of the polynomial 
$$
\frac{\partial^{k-1}}{\partial x^{k-1}}x^{2(k-1)}
$$
did not help me. Do you have any ideas? Thank you in advance.
 A: We start from
$$\sum_{q=1}^{n-1} {n\choose q}
\prod_{l=2}^q (4l-6) \prod_{l=2}^{n-q} (4l-6)
= \prod_{l=2}^n (4l-6).$$
This is
$$\sum_{q=1}^{n-1} {n\choose q}
2^{q-1} \prod_{l=2}^q (2l-3) 2^{n-q-1} \prod_{l=2}^{n-q} (2l-3)
= 2^{n-1} \prod_{l=2}^n (2l-3)$$
or 
$$\sum_{q=1}^{n-1} {n\choose q}
\prod_{l=2}^q (2l-3) \prod_{l=2}^{n-q} (2l-3)
= 2 \prod_{l=2}^n (2l-3)$$
or
$$\sum_{q=1}^{n-1} {n\choose q}
\frac{(2q-2)!}{2^{q-1} (q-1)!}
\frac{(2(n-q)-2)!}{2^{n-q-1} (n-q-1)!}
= 2 \frac{(2n-2)!}{2^{n-1} (n-1)!}$$
or
$$\sum_{q=1}^{n-1} {n\choose q}
\frac{(2q-2)!}{(q-1)!}
\frac{(2(n-q)-2)!}{(n-q-1)!}
= \frac{(2n-2)!}{(n-1)!}$$
Continuing we get 
$$n! \sum_{q=1}^{n-1} \frac{1}{q (n-q)}
{2q-2\choose q-1} {2n-2q-2\choose n-q-1} = \frac{(2n-2)!}{(n-1)!}$$
or
$$\sum_{q=1}^{n-1} \frac{1}{q (n-q)}
{2q-2\choose q-1} {2n-2q-2\choose n-q-1} = 
\frac{1}{n} {2n-2\choose n-1}.$$
Now recall the  generating function of the Catalan  numbers which says
that
$$\sum_{n\ge 0} \frac{1}{n+1} {2n\choose n} z^n
= \frac{1-\sqrt{1-4z}}{2z}.$$
This implies that
$$\sum_{n\ge 0} \frac{1}{n+1} {2n\choose n} z^{n+1}
= \frac{1-\sqrt{1-4z}}{2}$$
or
$$\sum_{n\ge 1} \frac{1}{n} {2n-2\choose n-1} z^{n}
= \frac{1-\sqrt{1-4z}}{2}.$$
Therefore the LHS of the identity is in fact
$$\sum_{q=1}^{n-1} [z^q] \frac{1-\sqrt{1-4z}}{2}
[z^{n-q}] \frac{1-\sqrt{1-4z}}{2}
\\ = \sum_{q=0}^{n} [z^q] \frac{1-\sqrt{1-4z}}{2}
[z^{n-q}] \frac{1-\sqrt{1-4z}}{2}
\\ = [z^n] \left(\frac{1-\sqrt{1-4z}}{2}\right)^2
= [z^n] \frac{1-2\sqrt{1-4z}+1-4z}{4}
\\ = [z^n] \frac{1-2z-\sqrt{1-4z}}{2}.$$
With $\sqrt{1-4z} = 1 - 2z - \cdots$ we get zero for $n\le 1$
and for $n\ge 2$ we find
$$[z^n] \frac{1-2z-\sqrt{1-4z}}{2}
= [z^n] \frac{1-\sqrt{1-4z}}{2} = 
\frac{1}{n} {2n-2\choose n-1}$$
as claimed.
