Legendre’s formula:

The exponent of p in the prime factorization of n! is

    \[ e_p(n!)= \sum_{r\ges 1} {\Big\lfloor \frac{n}{p^r} \Big\rfloor}\]

This is equivalent to:

    \[ n! = \prod_{p\les n} p^{\lfloor n/p \rfloor +\lfloor n / p^2 \rfloor +\lfloor n / p^3 \rfloor +\dots } \]

Explanation: Between 1 and n there are \lfloor n/p \rfloor numbers that are multiples of p : p,2p,3p,\dots,\lfloor n/p \rfloor p. (e.g. if p=3 and n=10, these are 3, 6, 9.) These each contribute a factor of p. Of those, there are \lfloor n/p^2 \rfloor multiples of p^2 that each contribute one more factor of p. (e.g. one more: 3^2=9. 4 factors in total.) The multiples of p^3 contribute one more factor, etc until \lfloor n/p^r \rfloor=0.
Example: Find the exponent of 5 in the prime factorization of 1000!.

    \[ \Big\lfloor \frac{1000}{5} \Big\rfloor +  \Big\lfloor \frac{1000}{5^2} \Big\rfloor +  \Big\lfloor \frac{1000}{5^3} \Big\rfloor +  \Big\lfloor \frac{1000}{5^4} \Big\rfloor = 200 + 40 + 8 + 1 = 249. \]