$$e^{-q} \sum_k \frac{q^k}{(k-1)!}$$
The next step is
$$qe^{-q} \sum_k \frac{q^k}{k!}$$
Where does the $q$ come from?
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They started by pulling out a $q$ to get $qe^{-q} \sum_k \frac{q^{k-1}}{(k-1)!}$, then re-indexed the sum, subbing $k$ for $k-1$. – Air Conditioner Aug 11 '19 at 19:14
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It's a change of variable. If you would write the limits, it should be more visible. Use $j$ instead of $k$ in the second expression, and they used $j=k+1$. The $q^j=qq^k$, and they moved one $q$ outside summation – Andrei Aug 11 '19 at 19:15
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You left out the summation limits, but it looks like they've been adjusted from, say, $\sum_{k=1}^n$ to $\sum_{k=0}^{n-1}$. Now, when you decrement the $k$ limits on the sum, you must increment $k$ in the expression to keep things balance. This is why $(k-1)!$ becomes $k!$. Likewise, $q^k$ becomes $q^{k+1}$ ... but that's $q\cdot q^k$, and that extra $q$ has been factored-out of the sum. – Blue Aug 11 '19 at 19:15
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Thankyou everyone!!!! I see it now! – Alex Tontodulie Taylor Aug 11 '19 at 19:36
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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.
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