{{tidy|use the equation template consistently}} {{begin-eqn}} {{eqn | l = \sin x       | r = x \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }       | c = Euler Formula for Sine Function }} {{eqn | l = \displaystyle \dfrac {\sin x} x       | r = \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} } = \paren {1 - \dfrac {x^2} {1^2 \pi^2} } \paren {1 - \dfrac {x^2} {2^2 \pi^2} } \paren {1 - \dfrac {x^2} {3^2 \pi^2} } \cdots }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sin x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots       | c = Power Series Expansion for Sine Function }} {{eqn | l = \displaystyle \dfrac {\sin x} x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n} } {\paren {2 n + 1}!} = 1 - \dfrac {x^2} {3!} + \dfrac {x^4} {5!} - \dfrac {x^6} {7!} + \cdots }} {{end-eqn}} Equating the product with the sum: {{begin-eqn}} {{eqn | l = \displaystyle \dfrac {\sin x} x       | r = \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }       | rr = = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n + 1}!} }} {{eqn | r = \paren {1 - \dfrac {x^2} {1 \pi^2} } x \paren {1 - \dfrac {x^2} {4 \pi^2} } x \paren {1 - \dfrac {x^2} {9 \pi^2} } x \cdots       | rr = = 1 - \dfrac {x^2} {3!} + \dfrac {x^4} {5!} - \dfrac {x^6} {7!} + \cdots }} {{end-eqn}} Each squared term in the product selected once: {{begin-eqn}} {{eqn | l = \displaystyle -\dfrac {x^2} {3!}       | r = -\dfrac {x^2} {\pi^2} \paren {1 + \dfrac 1 4 + \dfrac 1 9 + \dfrac 1 {16} + \cdots}       | c = Term raised to the second power used to calculate the Riemann Zeta Function of $2$ }} {{end-eqn}} Each unique combination of '''two''' squared terms in the product selected: :$\dfrac {x^4} {5!} = \dfrac {x^4} { \pi^4}   \paren {\paren 1 \paren {\dfrac 1 4} + \paren 1 \paren {\dfrac 1 9} + \paren 1 \paren {\dfrac 1 {16}} + \cdots + \paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren {\dfrac 1 4} \paren {\dfrac 1 {16}} + \cdots +  \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } }$ Term raised to the fourth power used to calculate Riemann Zeta Function of $4$ Each unique combination of '''three''' squared terms in the product selected: :$-\dfrac {x^6} {7!} = -\dfrac {x^6} {\pi^6}   \paren {\paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 {16} } + \cdots + \paren 1 \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots + \paren {\dfrac 1 4} \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots}$ Term raised to the sixth used power to calculate Riemann Zeta Function of $6$ {{begin-eqn}} {{eqn | l = \map \zeta 2       | r = \dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \cdots       | rr = = \dfrac {\pi^2} 6       | c = Basel Problem }} {{eqn | l = \paren {\map \zeta 2}^3       | r = \paren {\dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \cdots}^3       | rr = = \paren {\dfrac {\pi^2} 6}^3 = \dfrac {\pi^6} {216} }} {{end-eqn}} When we take the cube of a sum, we have: {{begin-eqn}} {{eqn | l = \paren {A + B + C + \cdots}^3       | r = \paren {A^3 + B^3 + C^3 + \cdots} + 3 \paren {A^2 B + A B^2 + A^2 C + A C^2 + B^2 C + B C^2 + \cdots} + 6 \paren {A B C + \cdots}       | c =  }} {{end-eqn}} Let $A = \dfrac 1 {1^2}, B = \dfrac 1 {2^2}, C = \dfrac 1 {3^2}, \cdots $ Then the {{LHS}} becomes: :$\paren {\paren {\dfrac 1 {1^2} } + \paren {\dfrac 1 {2^2} } + \paren {\dfrac 1 {3^2} } + \cdots}^3 = \paren {\map \zeta 2}^3$ and the first term on the {{RHS}} becomes: :$\paren {\paren {\dfrac 1 {1^2} }^3 + \paren {\dfrac 1 {2^2} }^3 + \paren {\dfrac 1 {3^2} }^3 + \cdots} = \map \zeta 6$ To make sense of the remaining two terms on the {{RHS}}, we need: :$\paren {AB + AC + BC + \cdots} = \dfrac {\pi^4} {5!} $ :$\paren {A + B + C + \cdots} = \dfrac {\pi^2} {3!} $ {{begin-eqn}} {{eqn | l = \paren {AB + AC + BC + \cdots}  \paren {A + B + C + \cdots}       | r = \paren {A^2B + AB^2 + A^2C + AC^2 + B^2C + BC^2 + \cdots} + 3\paren {ABC + \cdots}       | c =  }} {{eqn | l = 3 \paren {AB + AC + BC + \cdots} \paren {A + B + C + \cdots}       | r = 3 \paren {A^2B + AB^2 + A^2C + AC^2 + B^2C + BC^2 + \cdots} + 9 \paren {ABC + \cdots}       | c = We have 3 too many of the 'ABC' type (only need 6) - need to subtract 3 of these }} {{end-eqn}} Finally, we have: {{begin-eqn}} {{eqn | l = \paren {\map \zeta 2}^3       | r = \map \zeta 6  + 3 \dfrac {\pi^2} {3!} \dfrac {\pi^4} {5!} - 3  \dfrac {\pi^6} {7!}       | c = }} {{eqn | ll= \leadsto       | l = \map \zeta 6       | r = \paren {\map \zeta 2}^3  - 3  \dfrac {\pi^2} {3!} \dfrac {\pi^4} {5!} + 3  \dfrac {\pi^6} {7!}       | c = rearranging }} {{eqn | r = \dfrac {\pi^6} {216} - 21 \dfrac {\pi^6} {7!} + 3 \dfrac {\pi^6} {7!}       | c = simplifying }} {{eqn | r = 70 \dfrac {\pi^6} {3 x 7!} - 54 \dfrac {\pi^6} {3 x 7!}       | c = simplifying }} {{eqn | r = 16 \dfrac {\pi^6} {3 x 7!}       | c = simplifying }} {{eqn | r = \dfrac {\pi^6} {945}       | c = }} {{end-eqn}} {{qed}}