Guys, it depends on the maximum primes <= 100 (for considering (100!)^k). Why is that? Because, for a number a to be a factor of b, any prime number dividing into a should also divide into b. If p^k divides a, then p^k should also divide b, because otherwise, it would not divide properly, and remainders/fractions would result. If you take the equation:
10000! = (100!)^k
every prime factor (with powers) dividing (100!)^k should be dividing 10000! too. That means, if you consider some prime number p <= 100, p^k should divide 10000! too. Now, after some inspection, you can conclude that the formula for prime p's power in n! is "Floor(n/p) + Floor(n/(p^2)) + Floor(n/(p^3)) + ...".
If you try to understand powers of 2, in both parts of equation, what you get is, the following:
Notice that, for Floor(100/2) to Floor(100/64), exponents are proportionally present in 10000!. So, this means, for any prime p below 100, 10000/100 = 100 is a safe exponent. The problem is with the latter terms being present in 10000!, like Floor(10000/128) etc... Also, notice that these will be more (and of more value), if prime number p is small.
Since all primes dividing (100!)^k should also divide 10000!, we need to find a common exponent k, that works for all primes <= 100. Since the common exponent is minimum one, we need to find the prime <= 100, which results in a minimum exponent k, namely the largest prime, 97.
I think I got it. He segregated the categories of primes into ones which have the same power in 100!, and considered the maximum in each interval, as that is sufficient, good. Now, since the ratio matters than the actual numbers, we need to evaluate for all these 11 numbers, ok. Got it, great!