User blog comment:DrCeasium/new hyperfactorial array notation/@comment-27.130.114.45-20150725134508

n!k = n^(k)n-1^(k)13:45, July 25, 2015 (UTC)~ 3^(k)2^(k)1... n[1]= n!n n[2]= (n[1])[1])13:45, July 25, 2015 (UTC)... {n times} n[n]= (n[n-1])[n-1])... {n times} n[1,1]= n[n]