How does $e^{-q} \sum_k \frac{q^k}{(k-1)!}$ become $qe^{-q} \sum_k \frac{q^k}{k!}$?

Question

$$e^{-q} \sum_k \frac{q^k}{(k-1)!}$$

The next step is

$$qe^{-q} \sum_k \frac{q^k}{k!}$$

Where does the $q$ come from?

Answer 1

Looks like an index shift: $e^{-q} \sum_{k=1}^n \frac{q^k}{(k-1)!} = e^{-q} \sum_{k=0}^{n-1} \frac{q^{k+1}}{k!}$. Since $q^{k+1} = q^k \cdot q$, we know where the $q$ comes from. If you could write the limits, it would be easier to understand the anwsers to your question by the way.