Three. (Full marks 25 points) The product of $n$ consecutive natural numbers from 1 to $n$ is called the factorial of $n$, denoted as $n$! (For example, 5! $=5 \times 4 \times 3 \times 2 \times 1$). How many consecutive zeros are at the end of 1999! ? Explain your reasoning.

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Note: The original text uses "畍居" which seems to be a typo or an incorrect character. The correct term should be "reasoning" or "justification."
Three, consider the prime factorization of 1999!
$$
1999!=2^{\alpha_{1}} \cdot 3^{\sigma_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \ldots
$$

where the various $\alpha_{i}$ are non-negative integers. The number of trailing zeros in 1999! depends on $\alpha_{1}$ and $\alpha_{3}$, and is equal to the smaller of these two numbers. The exponent of 5 in the prime factorization is (where $[x]$ denotes the greatest integer less than or equal to $x$)
$$
\begin{aligned}
\alpha_{3} & =\left[\frac{1999}{5}\right]+\left[\frac{1999}{5^{2}}\right]+\left[\frac{1999}{5^{3}}\right]+\left[\frac{1999}{5^{4}}\right]+\cdots \\
& =399+79+15+3=496 .
\end{aligned}
$$

Similarly, $\alpha_{1}=1990>\alpha_{3}$ can be calculated, so 1999! has 496 trailing zeros.