Normalization of integral involving Laguerre polynomials

Question

The following integral involving Laguerre polynomials came up in a quantum mechanics problem I was working on: $$\int_{0}^{\infty} | {Akp^{l}e^{-p/2} L_{n-l-1}^{2l+1}}(p)|^{2} r^2dr$$ Where $p=2kr$ and $A$ is the normalization constant. According to my book, the normalization constant is $$A = \left[2k\tfrac{(n-l-1)!}{2n((n+l)!)^{3}}\right]^{1/2}$$ although I can't seem to get this. I first expanded the integral to get $$\int_{0}^{\infty} {A^{2}k^{2}p^{2l}e^{-p} [L_{n-l-1}^{2l+1}}(p)]^2r^2dr = \frac{1}{2k}A^{2}\int_{0}^{\infty} {p^{2l+2}e^{-p} [L_{n-l-1}^{2l+1}}(p)]^2dp$$ Multiplying each term of the recurrence relationship $$pL_{n}^{a}(p)=-(n+1)L_{n+1}^{a}(p)+(2n+a+1)L_{n}^{a}(p)-(n+a)L_{n-1}^{a}(p)$$ by $$p^{2l+1}e^{-p}L_{n-l-1}^{2l+1}(p)$$ and then integrating reduces all but the middle term to zero due to the orthogonality relationship: $$\int_{0}^{\infty} {p^{a}e^{-p}L_{n}^{a}}(p)L_{m}^{a}dp = \frac{(n+a)!}{n!}\delta_{n,m}$$ I then find that the integral is equal to $$\int_{0}^{\infty} {(pL_{n-l-1}^{2l+1}(p)) p^{2l+1}e^{-p}L_{n-l-1}^{2l+1}}(p)dp = \frac{(2n)(n+l)!}{(n-l-1)!}$$ Solving for $A$, I get $$A = \left[2k\tfrac{(n-l-1)!}{2n(n+l)!}\right]^{1/2}$$ This is notably lacking the cubed factorial in the denominator, and I'm at a loss as to where that would ever come up during integration. Can anyone spot any errors in my work?