Binary multiplication for negative numbers

Question

The question is about binary multiplication for negative numbers. Assume we want to multiply -5 * -3 so the result is +15.

1) In the first step, we have to use 2's complement for the inputs.

+5 = 0101    ->   -5 = 1011
+3 = 0011    ->   -3 = 1101

2) We follow the simple pencil-and-paper method and we have to note the sign extension. For the sake of clarity I put the signs extensions in []

                         1011
                       * 1101
               ----------------
   [1] [1] [1]  1   0    1    1
   [0] [0]  0   0   0    0 
   [1]  1   0   1   1
    1   0   1   1                 +
   ------------------------------
    c7  c6  c5  c4  c3   c2   c1

3) summing the columns show that

c1 = 1 + 0 + 0 + 0 = 1
c2 = 1 + 0 + 0 + 0 = 1
c3 = 0 + 0 + 1 + 0 = 1
c4 = 1 + 0 + 1 + 1 = 1   (carry 1 to c5)
c5 = 1(carry) + 1(sign) + 0 + 0 + 1 = 1 (carry 1 to c6)
c6 = 1(carry) + 1(sign) + 0(sign) + 1 + 0 = 1 (carry 1 to c7)
c7 = 1(carry) + 1(sign) + 0(sign) + 1(sign) + 1 = ???

Actually c7 = 100 but we have to represent only one digit in c7. So does that mean

c7 = 0 (carry 10)

?? Usually the final carry bit is only one digit. More than that, the final number is not equal to +15. Where did I make the mistake?

Answer 1

                         1011
                       * 1101
               ----------------
[1][1] [1] [1]  1   0    1    1
   [0] [0]  0   0   0    0 
[1][1]  1   0   1   1
[1] 1   0   1   1                 +
1   0   1   1
... 1   1
   ------------------------------
    c7  c6  c5  c4  c3   c2   c1

Does this clarify it?

If you keep going to the left, you will wind up carrying infinitely many bits. But that's as it should be. The actual answer on the right winds up as ...01111 when you include the 1011 entry ending in column 5. The more 1011 entries you include (ending at c6, c7, etc.) the more zeros you'll have at the start of your answer.

I don't know any textbook answer for where to stop going to the left, but you can at least see it conceptually from the above.