What is the sum of (n-1)+(n-2)+...+(n-k)? What is the sum of this series ?
$(n-1)+(n-2)+(n-3)+...+(n-k)$ 
$(n-1)+(n-2)+...+3+2+1 = \frac{n(n-1)}{2}$
So how can we find the sum from $n-1$ to $n-k$ ?

$$(n-1)+(n-2)\cdots(n-k)=\underbrace{n+n+\cdots +n}_{\text{$k$ copies}}-(1+2+\cdots k)=nk-\frac{k}{2}(k+1)$$
