Product to n of Product to Index

Theorem

$\ds \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j = \prod_{i \mathop = 0}^n {a_i}^{n + 2}$

Let:

$\ds P_1$

$:=$

$\ds \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j$

$\ds P_2$

$:=$

$\ds \prod_{i \mathop = 0}^n \prod_{j \mathop = i}^n a_i a_j$

Then:

$\ds P_1 P_2$

$=$

$\ds \paren {\prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j} \paren {\prod_{i \mathop = 0}^n \prod_{j \mathop = i}^n a_i a_j}$

$\ds $

$=$

$\ds \prod_{i \mathop = 0}^n \paren {\paren {\prod_{j \mathop = 0}^i a_i a_j} \paren {\prod_{j \mathop = i}^n a_i a_j} }$

$\ds $

$=$

$\ds \prod_{i \mathop = 0}^n \paren {\paren {\prod_{j \mathop = 0}^n a_j a_j} {a_i}^2}$

$\ds $

$=$

$\ds \prod_{i \mathop = 0}^n \paren { {a_i}^{n + 1} \paren {\prod_{j \mathop = 0}^n a_j} {a_i}^2}$

$\ds $

$=$

$\ds \paren {\prod_{i \mathop = 0}^n {a_i}^{n + 1} } \paren {\prod_{j \mathop = 0}^n \paren {\prod_{j \mathop = 0}^n a_j} {a_i}^2}$

$\ds $

$=$

$\ds \paren {\prod_{i \mathop = 0}^n {a_i}^{n + 1} } \paren {\prod_{j \mathop = 0}^n {a_i}^{n + 1} {a_i}^2}$

$\ds $

$=$

$\ds \prod_{i \mathop = 0}^n {a_i}^{2 n + 4}$

$\ds \leadsto \ \ $

$\ds P_1$

$=$

$\ds \prod_{i \mathop = 0}^n {a_i}^{n + 2}$

$\blacksquare$