Simplifying a nested sum
Hint:
$$\sum_{i=1}^n \sum_{j=i}^n (n-j) = \sum_{i=1}^n \sum_{j=0}^{n-i} j. $$
At this point, you can either use exact formulas, or notice that for all $d \geq 0$,
$$\sum_{k=1}^m k^d = \Theta(k^{d+1}).$$
(If you don't know what a $\Theta(\cdot)$ bound is, suffice it to say that it is a "tight" $O(\cdot)$ bound.)

After you handle the inner sum, you will need to use the same trick of replacing $i$ with $n-i$ in order to handle the outer sum.