A Dizzying Polynomial

Let $x=1$ , then:

${ P }_{ n }\left( 1 \right) =\prod _{ i=0 }^{ n }{ \sum _{ j=0 }^{ i }{ \prod _{ k=0 }^{ j }{ \sum _{ l=0 }^{ k }{ { 1 }^{ l } } } } } =\prod _{ i=0 }^{ n }{ \sum _{ j=0 }^{ i }{ \prod _{ k=0 }^{ j }{ \sum _{ l=0 }^{ k }{ { 1 } } } } } =\prod _{ i=0 }^{ n }{ \sum _{ j=0 }^{ i }{ \prod _{ k=0 }^{ j }{ \left( k+1 \right) } } } =\prod _{ i=0 }^{ n }{ \sum _{ j=0 }^{ i }{ \left( j+1 \right) ! } }$

Let's refine this expression a little bit:

${ P }_{ n }\left( 1 \right) =\prod _{ i=0 }^{ n }{ \sum _{ j=0 }^{ i }{ \left( j+1 \right) ! } } =\prod _{ i=0 }^{ n }{ \sum _{ j=1 }^{ i+1 }{ j! } } =\prod _{ i=1 }^{ n+1 }{ \sum _{ j=1 }^{ i }{ j! } }$

Now that we have a simpler task at hand we can start computing the wanted value:

$\frac { { P }_{ 5 }\left( 1 \right) }{ { P }_{ 3 }\left( 1 \right) { P }_{ 2 }\left( 1 \right) } =\frac { \prod _{ i=1 }^{ 6 }{ \sum _{ j=1 }^{ i }{ j! } } }{ \prod _{ i=1 }^{ 4 }{ \sum _{ j=1 }^{ i }{ j! } } \times \prod _{ i=1 }^{ 3 }{ \sum _{ j=1 }^{ i }{ j! } } } =\frac { \prod _{ i=5 }^{ 6 }{ \sum _{ j=1 }^{ i }{ j! } } }{ \prod _{ i=1 }^{ 3 }{ \sum _{ j=1 }^{ i }{ j! } } } =\\ \\ =\frac { \left( 1!+2!+3!+4!+5! \right) \times \left( 1!+2!+3!+4!+5!+6! \right) }{ \left( 1! \right) \times \left( 1!+2! \right) \times \left( 1!+2!+3! \right) } =\\ \\ =\frac { \left( 1+2+6+24+120 \right) \times \left( 1+2+6+24+120+720 \right) }{ 1\times \left( 1+2 \right) \times \left( 1+2+6 \right) } =\\ \\ =\frac { 153\times 873 }{ 1\times 3\times 9 } =\boxed { 4947 }$