Summing up decrementing geometric series?

Question

Is there any easy way of summing up,

$c,z \in R$

$z < 1, c < z $

$ k,n\in N$:

$$\large\sum_{k=0}^{\lfloor\frac{z}{c}\rfloor}\prod_{n=0}^{k}(z-nc)^n$$

I'm searching for a formula to sum up winstreaks in an online cardgame and want to account for matchmaking, which increases the difficulty after every win. Like the title says this seems similar to the partial sum of a geometric series, I haven't found any neat way to sum it up myself or online yet.

Answer 1

Ok, if c divides z, then you can turn around the product and sum to be

$$\large\sum_{k=0}^{\frac{z}{c}}\prod_{n=0}^{k}(nc)^{k-n}$$

I think ? I'm not 100% sure on that one, in my head it works. Which means you can come up with a upper/lower bound for any c, I think.