Making single product into double product I have this product: $\prod_{i > j} ( i^2 - j^2 ) $ and I am trying to turn it into either a double product with one for $j$ and the other for $i$ or keep as a single product and rearrange $j$ in terms of $i$. (Note the $i,j$ are for matrix indexes).
I first thought $\prod_{i = 1} ( i^2 - (i-1)^2 ) $ however testing this a few times shows that it does not work... can anyone help me? (The products are up to $n$)
Note I am trying to remove $i>j$ from my product by splitting into two or some other way
 A: You can write it as $\prod_{i=1}^n\prod_{j=1}^{i-1}(i^2-j^2)$ or $\prod_{j=1}^n\prod_{i=j+1}^n(i^2-j^2)$.
A: First way
$$
\eqalign{
  & P(n) = \prod\limits_{0\, \le \,j\, < \,i\, \le \,n}
 {\left( {i^{\,2}  - j^{\,2} } \right)}  = \prod\limits_{0\, \le \,j\, \le \,n - 1} {\prod\limits_{j + 1\, \le \,i\, \le \,n} {\left( {i^{\,2}  - j^{\,2} } \right)} }  =   \cr 
  &  = \prod\limits_{0\, \le \,j\, \le \,n - 1}
 {\prod\limits_{j + 1\, \le \,i\, \le \,n} {\left( {i + j} \right)\left( {i - j} \right)} }  =   \cr 
  &  = \prod\limits_{0\, \le \,j\, \le \,n - 1}
 {\prod\limits_{1\, \le \,i - j\, \le \,n - j}
 {\left( {i - j + 2j} \right)\left( {i - j} \right)} }  =   \cr 
  &  = \prod\limits_{0\, \le \,j\, \le \,n - 1}
 {\prod\limits_{1\, \le \,k\, \le \,n - j} {\left( {k + 2j} \right)k} }  =   \cr 
  &  = \prod\limits_{0\, \le \,j\, \le \,n - 1}
 {\left( {n - j} \right)!\left( {1 + 2j} \right)^{\,\overline {\,n - j\,} } }  =   \cr 
  &  = \quad  \cdots  \cr} 
$$
where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial
Second way
$$
\eqalign{
  & P(n) = \prod\limits_{0\, \le \,j\, < \,i\, \le \,n} {\left( {i^{\,2}  - j^{\,2} } \right)}  =   \cr 
  &  = \prod\limits_{1\, \le \,i\, \le \,n}
 {\prod\limits_{0 \le j\, \le \,i - 1} {\left( {i + j} \right)\left( {i - j} \right)} }  =   \cr 
  &  = \prod\limits_{1\, \le \,i\, \le \,n}
 {i^{\,\overline {\,i\,} } i^{\,\underline {\,i\,} } }
  = \prod\limits_{1\, \le \,i\, \le \,n} {i!{{\Gamma (2i)} \over {\Gamma (i)}}}  =   \cr 
  &  = \prod\limits_{1\, \le \,i\, \le \,n}
 {\Gamma (i + 1){{\Gamma (2i)} \over {\Gamma (i)}}}
  = \prod\limits_{1\, \le \,i\, \le \,n} {i\,\Gamma (2i)}  \cr} 
$$
And introducing the Duplication formula for the Gamma function
$$
\Gamma \left( {2\,z} \right) = {{2^{\,2\,z - 1} } \over {\sqrt \pi  }}\Gamma \left( z \right)
\Gamma \left( {z + 1/2} \right)
$$
we get
$$
\eqalign{
  & P(n) = \prod\limits_{1\, \le \,i\, \le \,n} {{{4^{\,i} } \over {2\sqrt \pi  }}
\Gamma (i + 1)\Gamma \left( {i + 1/2} \right)}  =   \cr 
  &  = \left( {{{2^{\,n} } \over {\sqrt \pi  }}} \right)^{\,n}
 \prod\limits_{1\, \le \,i\, \le \,n} {\Gamma (i + 1)\Gamma \left( {i + 1/2} \right)}  \cr} 
$$
