Show that k! divides the product of any k consecutive positive integers.

For example, 7.8.9.10.11 = 55440 is divisible by 5! = 120, with a quotient of 462.

We'll proceed by induction on the largest of the k consecutive integers. The theorem is that for 1 <= k <= n, the product of k consecutive integers, with n being the largest of them, is divisible by k!.

Let's do a few cases to get started (though only the case n = 1 is strictly necessary). For n = 1, the only choice for k is 1, and clearly 1 is a multiple of 1!. For n = 2, 2 is a multiple of 1!, and 1.2 is a multiple of 2!. For n = 3, 3 is a multiple of 1!, 2.3 is a multiple of 2!, and 1.2.3 is a multiple of 3!.

Assume that the theorem is true for any n = r. In other words, the product of k consecutive integers, with the largest of them equal to r, is divisible by k!, for 1 <= k <= r.

Now we'll show that the theorem is true for n = r + 1. If k is equal to 1, the theorem is trivially true, since any integer is a multiple of 1!. So, for k > 1, the product of k consecutive integers, with the largest of them being equal to r+1 is:

(r+1)r...(r-k+2)

= (r + 1 - k + k)r(r-1)...(r-k+2)

= r(r-1)...(r-k+1) + kr(r-1)...(r-k+2)

The first term on the right hand side is the product of k consecutive integers with the largest of them being equal to r, and so this term is divisible by k!, by the inductive hypothesis.

The second term on the right hand side contains the factor r(r-1)...(r-k+2), which is the product of k-1 consecutive integer with the largest of them begin r, and so by the inductive hypothesis, this factor is divisible by (k-1)!. Hence the second term on the right hand side, being k times this factor, is divisible by k!.

As a result the entire right hand side is divisible by k!, and the theorem is proven.