We recall that $n!=1 \times 2 \times 3 \times \cdots \times n$.

How many zeros are there at the end of 100! ? and at the end of 1000! ?
There are more multiples of 2 than multiples of 5 (and more multiples of $2^{2}$ than multiples of $5^{5}$, etc.) in the numbers between 1 and 100 (or 1000). Therefore, it suffices to evaluate the power of 5 in the prime factorization of 100! or 1000!. The number of multiples of 5 between 1 and 1000 is $1000 / 5=200$, the number of multiples of $5^{2}$ is $1000 / (5^{2})=40$, the number of multiples of $5^{3}$ is $1000 / 5^{3}=8$, and the number of multiples of $5^{4}$ is the integer part of $1000 / 5^{4}$, which is 1. The power of 5 in 1000! is therefore:

$$
200+40+8+1=249
$$

Similarly, the power of 5 in 100! is:

$$
20+4=24
$$