证明连续N个正整数的乘积能被N!整除

车五进二

Let p be any prime number between 1 and n.In n!, there are [n/p]
consecutive multiples p, where [x] is the greatest integer less
than or equal to x.

Therefore the number of factors of p in n! is [n/p] + [n/(p^2)] +
[n/(p^3)]...

For N consecutive numbers, by the pigeon hole theorem, there are
at least [n/p] consecutive multiples of p, therefore the number
of factors of p in the product is at least [n/p] + [n/(p^2)] +
[n/(p^3)]..., hence divisible by the factors of p in n!.

Since this applies to all the prime factors of n!, therefore
K(K+1)...(K+N-1) is divisible by n!.