I think I figured this one out. The continued product of each ## j ## from ## 1 ## to ## l+1 ## is the ## ((l+1)!)^2 ##. The 2's in the numerator actually can be divided in the denominator. The sequence of 1*3*5*7...(squared) in the denominator takes on two forms depending whether the last term is ## (2l+1)/2 ## or includes the ## (l+\frac{3}{2}) ## term, thereby the continued product of ## (2j-1)/2 ## and ## (2j+1)/2 ## in the denominator. The ## \sqrt{\pi} ## in the denominator gets squared and moved to the numerator of the other side, and the ## \frac{1}{2} ## is a simple algebraic term because ## \frac{1}{(2l+3)/2}=\frac{2}{2l+3} ##.