How many 0s are there at the end of the number 100!? Recall that $n!=1 \times 2 \times \cdots \times n$.

保留源文本的换行和格式，这里的翻译结果已经符合要求。
The problem comes down to counting the number of factors of 5 in the product $1 \times 2 \times 3 \times \cdots \times 100$ (since there are as many zeros as there are factors of 10, and clearly there are more factors of 2 than factors of 5). Every 5 numbers contribute one factor of 5, but every $5^{2}=25$ contributes a second one (there are never 3 in a single number because that number would be greater than $5^{3}=125$). Therefore, there are $\frac{100}{5}=20$ factors of 5 and $\frac{100}{25}=4$ factors of 25. Thus, there are $20+4=24$ zeros at the end of $100!$!