Starting with n numbers a(1), a(2), ..., a(n) we will always end up with (a(1)+1) ( a(2)+1)...(a(n)+1) -1. The reason is that ab+a+b = (a+1)(b+1) – 1 One way to prove the above statement is by induction on n: Lets say in the step right before the last step we are left with two numbers a and b. We know a and b are results of this operation on k and n-k numbers for some k < n. So by induction hypothesis a = (a(1)+1)...(a(k)+1) -1 and b =(a(k+1)+1)...(a(n)+1) -1 Calculating (a+1)(b+1)-1 proves the statement for n, which shows the statement is true by induction on n.