An identity involving the Pochhammer symbol I need help proving the following identity:
$$\frac{(6n)!}{(3n)!} = 1728^n \left(\frac{1}{6}\right)_n \left(\frac{1}{2}\right)_n \left(\frac{5}{6}\right)_n.$$
Here,
$$(a)_n = a(a + 1)(a + 2) \cdots (a + n - 1), \quad n > 1, \quad (a)_0 = 1,$$
is the Pochhammer symbol. I do not really know how one converts expressions involving factorials to products of the Pochhammer symbols. Is there a general procedure? Any help would be appreciated.

$$1728^n \left(\frac{1}{6}\right)_n \left(\frac{1}{2}\right)_n \left(\frac{5}{6}\right)_n=24^n\prod_{k=0}^{n-1}{(6k+1)(2k+1)(6k+5)}=\\=\frac{4^n}{3^n n!}\prod_{k=0}^{n-1}{(6k+1)(6k+3)(6k+5)(6k+6)}=\frac{\prod_{k=0}^{n-1}{(6k+1)(6k+2)(6k+3)(6k+4)(6k+5)(6k+6)}}{\prod_{k=0}^{n-1}{(3k+1)(3k+2)(3k+3)}}=\frac{(6n)!}{(3n)!}$$
