with how many zeros it ends I have to calculate with how many zeros 1000! ends. This is wat I did, but I am not sure whether its good: I think I have to calculate how many times the product 10 is in 1000!. I found out the factor 10 is 249 times in 1000! using the fact that $s_p(n!)=\sum_{r=1}^{\infty}\lfloor \frac{n}{p^r}\rfloor$. So I think the answer should be that it ends with 249 zeros.
Is this correct/are there other ways to do this? If not how should I do it then?

Also, your sum doesn't have to go to infinity. As you no doubt noticed, the sum itself is finite and the terms after a first few finite terms are all zero. We actually know exactly where we can truncate the sum. We will stop when we hit the largest power of $p$ dividing $n$ which can easily be found by $\left\lfloor \frac{log(n)}{log(p)} \right\rfloor$ because after that the fraction will be less than one so taking its floor will always give zero. So your sum can be written as
$$\sum_{r=1}^{\left\lfloor\frac{\log(n)}{\log(p)}\right\rfloor} \left\lfloor\frac{n}{p^r}\right\rfloor$$
