Tricky induction proof: pile of stones split into n groups So I understand setting up this proof and the basis step just fine, however, it is the induction step where I am completely lost. I went and asked the math tutors at my school, the the tutor that spoke with me even had a tough time with this problem... it is supposed to be tricky, any help would be appreciated, thank you!
Suppose you begin with a pile of n stones (n ≥ 2) and split this pile into n piles of one stone each by
successively splitting a pile of stones into two smaller piles. Each time you split a pile you multiply the
number of stones in each of the two smaller piles you form, so that if these piles have p and q stones
in them, respectively, you compute pq. Show that no matter how you split the piles (eventually into n
piles of one stone each), the sum of the products computed at each step equals n(n − 1)/2. (Hint: use
strong induction on n.)
My attempt at this question makes no sense, and I just couldn't make sense of the tutors words in the time I had with her. Again, any help would be much appreciated!

At the first step you have piles of size $a$ and $n-a$, and write down the product $a(n-a)$.  Now you will handle the pile of size $a$; by the strong inductive hypothesis the sum of all the products you write down (as you break this sub-pile down all the way) will be $\frac{a(a-1)}{2}$.  Next, you handle the pile of size $n-a$; by the strong inductive hypothesis the sum of all the products you write down will be $\frac{(n-a)(n-a-1)}{2}$.  The total will be
$$a(n-a)+\frac{a(a-1)}{2}+\frac{(n-a)(n-a-1)}{2}$$
Adding these and simplifying yields $\frac{n(n-1)}{2}$.
The only thing left is the base case of $n=1$, where there is nothing to do, and you write down nothing.  This equals $\frac{1(0)}{2}$, so all is well.
