Algebra question $n\sum_{k=0}^n {n-1 \choose k-1} p^{k}q^{n-k} = np\sum_{k=1}^n {n-1 \choose k-1} p^{k-1}q^{n-k}$

Question

In a proof in my textbook one step goes from ... $$n\sum_{k=0}^n {n-1 \choose k-1} p^{k}q^{n-k} = np\sum_{k=1}^n {n-1 \choose k-1} p^{k-1}q^{n-k}$$

I understand that you can take the $p$ out because $p^{k}$ changed to $p^{k-1}$. But, I don't understand how k can go from 0 to 1.

Answer 1

HINT What is $\dbinom{n-1}{-1}$, which corresponds to the $k=0$ term?