An algebra problem by Nuttapat Bk

$\sum _{ i=1 }^{ n }{ (i+1)!{ i }^{ 2 }\qquad } \\ =\quad \sum _{ i=1 }^{ n }{ (i+1)!({ i }^{ 2 }+5i+6)-\sum _{ i=1 }^{ n }{ (i+1)!(5i+5)-\sum _{ i=1 }^{ n }{ (i+1)! } } } \\ =\quad \sum _{ i=1 }^{ n }{ (i+1)!({ i }+2)(i+3)-\quad 5\sum _{ i=1 }^{ n }{ (i+1)!(i+1)-\sum _{ i=1 }^{ n }{ (i+1)! } } } \\ =\quad \sum _{ i=1 }^{ n }{ (i+3)! } \quad -\quad \sum _{ i=1 }^{ n }{ (i+1)! } \quad -\quad 5\sum _{ i=1 }^{ n }{ (i+1)!(i+1) } \\ \\ We\quad will\quad find\quad the\quad value\quad of\quad \sum _{ i=1 }^{ n }{ (i+1)!(i+1) } \quad first;\\ \sum _{ i=1 }^{ n }{ (i+1)!(i+1) } =\sum _{ i=1 }^{ n }{ (i+1)!(i+2-1) } \\ =\quad \sum _{ i=1 }^{ n }{ (i+1)!(i+2) } \quad -\quad \sum _{ i=1 }^{ n }{ (i+1)!(1) } \\ =\quad \sum _{ i=1 }^{ n }{ (i+1)!(i+2) } \quad -\quad \sum _{ i=1 }^{ n }{ (i+1)! } \quad =\quad \sum _{ i=1 }^{ n }{ (i+2)!\quad -\quad \sum _{ i=1 }^{ n }{ (i+1)! } } \\ =\quad 3!+4!+5!+...+n!+(n+1)!+(n+2)!\quad -\quad 2!-3!-4!-5!-...-m!-(n+1)!\\ =\quad (n+2)!-2\\ \\ Now\quad we\quad find\quad the\quad value\quad of\quad \sum _{ i=1 }^{ n }{ (i+3)!\quad } -\quad \sum _{ i=1 }^{ n }{ (i+1)! } ;\\ =\quad 4!+5!+6!+...+n!+(n+1)!+(n+2)!+(n+3)!\quad -\quad 2!-3!-4!-5!-6!-...-n!-(n+1)!\\ =\quad (n+2)!+(n+3(!\quad -2!-3!\\ =\quad (n+2)!+(n+3)!\quad -\quad 8\\ \\ Therefore\quad \sum _{ i=1 }^{ n }{ (i+1)!{ i }^{ 2 } } \\ =\quad \sum _{ i=1 }^{ n }{ (i+3)! } \quad -\quad \sum _{ i=1 }^{ n }{ (i+1)! } \quad -\quad 5\sum _{ i=1 }^{ n }{ (i+1)!(i+1) } \\ =\quad (n+2)!+(n+3)!\quad -\quad 8\quad -\quad 5\left[ (n+2)!-2 \right] \\ =\quad (n+2)!(1+n+3-5)\quad +2\\ =\quad (n+2)!(n-1)\quad +\quad 2\\ \\ 2!{ 1 }^{ 2 }\quad +\quad ...\quad +\quad 601!{ 600 }^{ 2 }\\ =\quad \sum _{ i=1 }^{ 600 }{ (i+1)!{ i }^{ 2 } } =\quad (602!)599\quad +\quad 2\\$