Theorem: Riemann Zeta Function of 6

The Riemann zeta function of $6$ is given by:
{{begin-eqn}}
{{eqn | l = \map \zeta 6
      | r = \dfrac 1 {1^6} +  \dfrac 1 {2^6} +  \dfrac 1 {3^6} +  \dfrac 1 {4^6} + \cdots
      | c = 
}}
{{eqn | r = \dfrac {\pi^6} {945}
      | c = 
}}
{{eqn | o = \approx
      | r = 1 \cdotp 01734 \, 3 \ldots
      | c = 
}}
{{end-eqn}}

Proof:
{{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 | r = x \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 = x\sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n} } {\paren {2 n + 1}!}
      | c = Power Series Expansion for Sine Function
}}
{{eqn | r = x \paren {1 - \dfrac {x^2} {3!} + \dfrac {x^4} {5!} - \dfrac {x^6} {7!} + \cdots }
}}
{{end-eqn}}
Dividing out the x factor on both sides and equating the product with the sum, we have:
{{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }
      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n + 1}!}
}}
{{eqn | l = \paren {1 - \dfrac {x^2} {1 \pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \cdots
      | r = \paren {1 - \dfrac {x^2} {3!} + \dfrac {x^4} {5!} - \dfrac {x^6} {7!} + \cdots }
}}
{{end-eqn}}
Equating the $x^2$ term on both sides of the equation, we obtain the value of the sum of the individual terms in the Basel Problem:
{{begin-eqn}}
{{eqn | l = -\dfrac {x^2} {3!}
      | r = -\dfrac {x^2} {\pi^2} \paren {1 + \dfrac 1 4 + \dfrac 1 9 + \dfrac 1 {16} + \cdots}
      | c = Basel Problem
}}
{{eqn | l = \dfrac {\pi^2} 6
      | r = \paren {1 + \dfrac 1 4 + \dfrac 1 9 + \dfrac 1 {16} + \cdots}
      | c = 
}}
{{end-eqn}}
Equating the $x^4$ term on both sides of the equation, we obtain the value of the sum of the product of every unique combination of $2$ terms from the Basel Problem:
{{begin-eqn}}
{{eqn | l = \dfrac {x^4} {5!}
      | r = \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} } + \cdots }
      | c = 
}}
{{eqn | l = \dfrac {\pi^4} {120 }
      | r = \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} } + \cdots }
      | c = 
}}
{{end-eqn}}
Equating the $x^6$ term on both sides of the equation, we obtain the value of the sum of the product of every unique combination of $3$ terms from the Basel Problem:
{{begin-eqn}}
{{eqn | l = -\dfrac {x^6} {7!}
      | r = -\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}
      | c = 
}}
{{eqn | l = \dfrac {\pi^6} {7!}
      | r = \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}
      | c = 
}}
{{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 = 
}}
{{eqn | l = \paren {\text {Cube of Sum } }
      | r = \paren {\text {Sum of Cubes } } + 3 \paren { \text {Product of 3 Terms, with one term a perfect square (Every Combination) } } + 6 \paren { \text {Product of 3 Unique Terms (Every Combination) } }
      | c = 
}}
{{end-eqn}}
Let $A = \dfrac 1 {1^2}, B = \dfrac 1 {2^2}, C = \dfrac 1 {3^2}, \cdots $
Then the {{LHS}} (Cube of Sum) becomes:
{{begin-eqn}}
{{eqn | l = \paren {\paren {\dfrac 1 {1^2} } + \paren {\dfrac 1 {2^2} } + \paren {\dfrac 1 {3^2} } + \cdots}^3
      | r = \paren {\map \zeta 2}^3
      | c = 
}}
{{end-eqn}}
and the first term on the {{RHS}} (Sum of Cubes) becomes:
{{begin-eqn}}
{{eqn | l = \paren {\paren {\dfrac 1 {1^2} }^3 + \paren {\dfrac 1 {2^2} }^3 + \paren {\dfrac 1 {3^2} }^3 + \cdots}
      | r = \map \zeta 6
      | c = 
}}
{{end-eqn}}
To obtain the remaining two terms on the {{RHS}}, we have:
{{begin-eqn}}
{{eqn | l = \paren {AB + AC + BC + \cdots}
      | r = \dfrac {\pi^4} {5!}
      | c = From the $x^4$ term above: Product of $2$ Unique terms
}}
{{eqn | l = \paren {A + B + C + \cdots}
      | r = \dfrac {\pi^2} {3!}
      | c = From the $x^2$ term above: Each term once
}}
{{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 \paren { 7!} } - 54 \dfrac {\pi^6} {3 \paren {7!} }
      | c = simplifying
}}
{{eqn | r = 16 \dfrac {\pi^6} {3 \paren {7!} }
      | c = simplifying
}}
{{eqn | r = \dfrac {\pi^6} {945}
      | c =
}}
{{end-eqn}}
{{qed}}