1000 Zeros

The number of trailing zeroes in the expansion of $n!$ is given by $\displaystyle \sum_{i=1}^{k} \left \lfloor \frac {n}{5^i} \right \rfloor$ where $k$ must be chosen such that $5^{k+1}>n$ .

Okay, there are 1000 ÷ 5 = 200 multiples of 5 between 1 and 1000. The next power of 5, namely 25, has 1000 ÷ 25 = 40 multiples between 1 and 1000. The next power of 5, namely 125, will also fit in the expansion, and has 1000 ÷ 125 = 8 multiples between 1 and 1000. The next power of 5, namely 625, also fits in the expansion, and has 1000 ÷ 625 = 1.6... um, okay, 625 has 1 multiple between 1 and 1000. (I don't care about the 0.6 "multiples", only the one full multiple, so I truncate my division down to a whole number.)

In total, I have 200 + 40 + 8 + 1 = 249 copies of the factor 5 in the expansion, and thus:

249 trailing zeroes in the expansion of 1000!