for n in io.lines()do a={1}m=2 c=0 for i=2,n do k=1 while c>0 or k<m do t=c+i*(a[k]or 0)c=t/10-t/10%1 a[k]=t%10 k=k+1 end m=k end print(table.concat(a):reverse())end
Note that non-ascii characters in the above source code will be escaped (such as \x9f).