What's his last name? Of the Southern aisles

Let the list $A = {1,2 ...... c,b,a }$ ( c , b and a are the last three numbers )

$STEP - 1 :$ Let the two numbers be a and b

$STEP - 2 :$ Sum = a + b product = ab

$STEP - 3 :$ adding numbers in step 2 we get $a + b + ab = (a+1)(b+1) - 1$ This realization is crucial (You might want to remember this as a beautiful binary operation)

$STEP -4 :$ adding number to list and eliminating a and b from the set we have $A = {1,2, ......... c , (a+1)(b+1) - 1}$

Now if we pick up the pair $(c, (a+1)(b+1) - 1)$ Letting $A = (a+1)(b+1) - 1$ we see that the pair ends up being $(A+1)(c+1) - 1$ OR equivalently $(a+1)(b+1)(c+1) - 1$ .... And so on

So at the end it is easy to see that WE HAVE A VERY BIG NUMBER

$(10^{12} + 1)! - 1$ A TRILLION FACTORIAL! it has a very large number of trailing zeroes ........ minus one yields a number with a greater number of trailing 9's

Hence the answer is $\boxed{999999999}$

NOTE : The identity in step 3 is folklore and employed in many problems of the recursive kind ....... try this

Define a set $A$ as the set of the first one trillion positive integers. Then, define set $B$ as the set containing every element in set $A$ plus one (i.e. set $B$ initially contains the positive integers $2,3,4,5,6,7 ... ... ... , 1 000 000 000, 1 000 000 001$ ).

When we apply the given operation on two integers from set $A$ , we are replacing the positive integers $x$ and $y$ with $x + y + xy$ . So, the operation replaces the positive integers $x + 1$ and $y + 1$ with $xy + x + y + 1 = (x+1)(y+1)$ in set $B$ .

Therefore, in set $B$ , the given operation is simply replacing two numbers with their product. So, when only one number is left, the number will be $2 \times 3 \times 4 \times 5 ... ... ... \times 1 000 000 000 \times 1 000 000 001$ This number obviously has at least 9 zeros at the end.

The final element in set $A$ will be $2 \times 3 \times 4 \times 5 ... ... ... \times 1 000 000 000 \times 1 000 000 001 - 1$ The last nine digits of this number are obviously $\boxed{ 999 999 999 }$