Among other similarities, why is that in every $S_{ a }=\sum _{ j=1 }^{ n }{ { j }^{ a } }$ the denominator is the product of the sum of the coefficients of powers of n of individual factors?
$ S_{ 1 }=\sum _{ j=1 }^{ n }{ j } =\left( \frac { n }{ 1 } \right) \cdot \left( \frac { n+1 }{ 2 } \right) \quad \\ Here\quad 1\quad and\quad 1+1=2\\ S_{ 2 }=\sum _{ j=1 }^{ n }{ { j }^{ 2 } } =\frac { n\left( n+1 \right) }{ 2 } \cdot \frac { \left( 2n+1 \right) }{ 3 } \\ Here\quad 2+1=3\\ S_{ 3 }=\sum _{ j=1 }^{ n }{ { j }^{ 3 } } =\left( \frac { n\left( n+1 \right) }{ 2 } \right) ^{ 2 }\\ Here\quad { \left( 1+1 \right) }^{ 2 }=4\\ S_{ 4 }=\sum _{ j=1 }^{ n }{ { j }^{ 4 } } =\frac { n\left( n+1 \right) }{ 2 } \cdot \frac { \left( 2n+1 \right) }{ 3 } \cdot \frac { \left( 3n^{ 2 }+3n-1 \right) }{ 5 } \\ Here\quad 3+3-1=5\\ S_{ 5 }=\sum _{ j=1 }^{ n }{ { j }^{ 5 } } =\left( \frac { n\left( n+1 \right) }{ 2 } \right) ^{ 2 }\cdot \frac { \left( 2n^{ 2 }+2n-1 \right) }{ 3 } \\ Here\quad 2+2-1=3\\ S_{ 6 }=\sum _{ j=1 }^{ n }{ { j }^{ 6 } } =\frac { n\left( n+1 \right) }{ 2 } \cdot \frac { \left( 2n+1 \right) }{ 3 } \cdot \frac { \left( 3n^{ 4 }+6n^{ 3 }-3n+1 \right) }{ 7 } \\ Here\quad 3+6-3+1=7\\ S_{ 7 }=\sum _{ j=1 }^{ n }{ { j }^{ 7 } } =\left( \frac { n\left( n+1 \right) }{ 2 } \right) ^{ 2 }\cdot \frac { \left( 3n^{ 4 }+6n^{ 3 }-n^{ 2 }-4n+2 \right) }{ 6 } \\ Here\quad 3+6-1-4+2=6 $
I have checked till $a=7$ and it happens invariably. Are there any explanation for this?
(or the nature is providing us a fantastic mnemonic in learning these formula?)
