Here is one solution:
# Caculate the number of trailing zeros in n! using de Polignac's formula:
# nzeros = sum_i floor(n/5^i)
import math
n = 126
nzeros = 0
term = n
while True:
term //= 5
if term == 0:
break
nzeros += term
print('{:d}! ends in {:d} zeros.'.format(n, nzeros))
print('{:d}! = {:d}'.format(n, math.factorial(n)))
For this example value of n the output is:
126! ends in 31 zeros.
126! = 23721732428800468856771473051394170805702085973808045661837377170052497697783313457227249544076486314839447086187187275319400401837013955325179315652376928996065123321190898603130880000000000000000000000000000000