We are to find the maximum value of k for which 2014^k divides (1!*2!*3!*...*2014!).
In that product of consecutive factorials, the smallest prime factor of 2014 will occur the most often, and the largest prime factor will occur least often. So the largest value of k for which 2014^k divides that product of factorials is the number of factors of 53 in that product of factorials.
The 1st set of 52 factorials -- 1! through 52! -- contain no factors of 53.
The 2nd set of 53 factorials -- 53! through 105! -- each contain 1 factor of 53 --> 53 factors of 53.
The 3rd set of 53 factorials -- 106! through 158! -- each contain 2 factors of 53 --> 53*2 = 106 factors of 53.
The 4th set of 53 factorials -- 159! through 211! -- each contain 3 factors of 53 --> 53*3 = 159 factors of 53.
...
...
The 37th set of 53 factorials -- 1961! through 2013! -- each contain 37 factors of 53 --> 53*37 = 1961 factors of 53.
The last factorial -- 2014! -- contains 38 factors of 53.
The total number of factors of 53 in the product of the consecutive factorials is
53(1+2+3+...+37)+38 = 37297
ANSWER: The product of factorials is divisible by 2014^37297 but not by 2014^37298.
