I seem to be stuck on the moment generating function of a geometric distribution.

so far

$\displaystyle m(t) = \sum_{y=0}^{\infty} e^{ty}p(y) = \sum_{y=0}^{n} e^{ty} pq^{y-1} = p \sum_{y=0}^{n} e^{ty} q^{y-1}$

how do you go from $\displaystyle p \sum_{y=0}^{n} e^{ty} q^{y-1} \ \ \mbox{to} \ \ p \sum_{y=0}^{n} (qe^t)^y$ where those the -1 in $\displaystyle p \sum_{y=0}^{n} e^{ty} q^{y {\color{red}-1}}$ go?