A number theory problem by K. J. W.

$Note\quad that\quad n!+(n+1)!+(n+2)!\\ \\ =\quad n!+n!(n+1)+n!(n+1)(n+2)\\ =\quad n!(1+n+1+{ n }^{ 2 }+3n+2)\\ =\quad n!({ n }^{ 2 }+4n+4)\\ =\quad n!{ (n+2) }^{ 2 }\\ \\ Therefore\quad n!\quad must\quad be\quad a\quad square\quad number.\\ \\ Note\quad that\quad if\quad p!\quad \le \quad x!\quad <\quad (2p)!\quad where\quad p\quad is\quad a\quad prime\quad number,\\ x!\quad is\quad divisible\quad by\quad p\quad but\quad not\quad by\quad { p }^{ 2 }.\quad Thus\quad x!\quad cannot\quad be\quad a\quad square\quad number.\\ \\ For\quad the\quad case\quad p=53,\quad x\quad can\quad be\quad any\quad number\quad from\quad 53\quad to\quad 100.\quad Thus\quad n\quad cannot\quad be\quad from\quad 53\quad to\quad 100.\\ \\ For\quad the\quad case\quad p=29,\quad x\quad can\quad be\quad any\quad number\quad from\quad 29\quad to\quad 52.\quad Thus\quad n\quad cannot\quad be\quad from\quad 29\quad to\quad 100.\\ \\ For\quad the\quad case\quad p=17,\quad x\quad can\quad be\quad any\quad number\quad from\quad 17\quad to\quad 28.\quad Thus\quad n\quad cannot\quad be\quad from\quad 17\quad to\quad 100.\\ \\ For\quad the\quad case\quad p=11,\quad x\quad can\quad be\quad any\quad number\quad from\quad 11\quad to\quad 16.\quad Thus\quad n\quad cannot\quad be\quad from\quad 11\quad to\quad 100.\\ \\ For\quad the\quad case\quad p=7,\quad x\quad can\quad be\quad any\quad number\quad from\quad 7\quad to\quad 10.\quad Thus\quad n\quad cannot\quad be\quad from\quad 7\quad to\quad 100.\\ \\ We\quad are\quad left\quad with\quad n=1,2,3,4,5,6\quad and\quad it\quad can\quad be\quad easily\quad checked\quad that\quad of\quad these,\quad only\quad n=1\quad satisfies\quad the\quad above\quad conditions.\\ \\ Therefore\quad the\quad largest\quad (and\quad only)\quad solution\quad is\quad n=1.\\$