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?

Actually, what you want to  count is the number of $5$'s. If you count how many multiples of $5$ show up (multiples of $25$ should be counted twice, and multiples of $125$ should count for three times), you will get the number of $5$'s in the prime factorization of $1000!$
Similarly, you can count the number of $2$'s in the prime factorization of $1000!$. This number will be much greater.
Since $10=2\times 5$, for each $5$ and $2$ in the prime factorization, you get a power of $10$, ie. a $0$ at the end of the number. So the number of $0$'s will be whichever is less, the number of $5$'s or the number of $2$'s.
