OK, so let me explain what the formula shows.

Let's consider the possible routes by which the attacker can go from the first territory to the second. He has to leave behind 1 piece, so there will be A[T_A]-1 attacking pieces against the D[T_1] defending pieces on the first territory he attacks. Given that we are calculating the probability that the attacker wins in his overall conquest, then the different outcomes by which victory might be possible are that: he has all A[T_A]-1 pieces left at the end, after the battle; he has A[T_A]-2 pieces left at the end; etc., all the way down to having only 1 piece left at the end (he cannot have 0 because then he hasn't won). Then the probability that he has a certain number, k, of pieces remaining, given the initial battle was A[T_A]-1 vs D[T_1], is simply the expectation from a single (A,D) battle that is calculated as before.

Now given that he had k pieces remaining, in the next battle, he will be able to fight using only k-1 pieces, against D[T_2] pieces. In this battle, again, any outcome from his ending up with k-1 pieces to his ending up with just 1 piece is possible. The probability of each of these is just the probability of his having a certain number of pieces remaining after a battle of (A,D), as before.

So now there is a pattern. My plan is simply to multiply the probability of the attacker having a pieces remaining after the first battle, b after the second, c after the third, together; and then sum over all values of the number of pieces remaining from each battle that are possible.

Maybe the best thing to request from you is that you could write out the function for a 2-series battle? i.e. A[T_A] on territory A, attacks territory 1 with D[T_1] defenders, and then territory 2 with D[T_2] defenders. What is the function for the probability the attacker will win both, in terms of A[T_A], D[T_1], D[T_2] and the single battle-probabilities we worked out before?