Yeah I got it. That trick works when the n in n x 9 is a whole number less or equal to 10. For n is greater than 10 it does not necessarily work. For example, when n=11 it fails since 11 x 9= 99 and 9 + 9 =18. So then the question is what kind of property do the sum of digits satisfy when we consider all n? Well, apparently you can prove that the sum of the digits is always divisible by 9. So in the case of 11 x 9 = 99 since 9 + 9 =18 it is easy to see that it is divisble by 9. Now let's try 12 x 9 =108 so 1+0 + 8 =9 so that is definitely divisble by 9, for 12345678901 x 9 = 111111110109 we have that 1+1+1+1+1+1+1+1+0+1+0+9=18 and that is divisble by 9 :D
(I thought that -> 9 x 9 = 81 = (8+1)x(8+1) was just a nice way of writing this property. Since everything is one equation. It also works for the others for example 9 x 8 = 72 = (7+2)x(7+2) :D )