The first thing you should notice is that every instance of a multiple
of 10 in the factorial expansion will yield a trailing zero. Next,
notice that multiples of 5 and 2 will yield trailing zeroes, including
the same yielded by the multiples of 10.

In the factorization of 1000000, there are far more 2s than there are
5s. Thus, in order to count the number of trailing zeroes, we need
only count the number of 5s that occur as factors.

This can be done simply with:
floor(1000000/5) + floor(1000000/5^2) + floor(1000000/5^3) +
floor(1000000/5^4) + floor(1000000/5^5) + floor(1000000/5^6) +
floor(1000000/5^7) + floor(1000000/5^8)

(And since 5^9 = 1953125 > 1000000, no further terms are needed past 5^8.)

The result:
200000 + 40000 + 8000 + 1600 + 320 + 64 + 12 + 2 = 249998

Two other related problems from Project Euler can be found here
and here.