Looking at David M. Burton's Elementary Number Theory the last few days and I've found an exercise where you are to show:
- $2$ divides $n(n+1)$
- $6$ divides $n(n+1)(n+2)$
- $24$ divides $n(n+1)(n+2)(n+3)$
- $120$ divides $n(n+1)(n+2)(n+3)(n+4)$
but the book does not go on to investigate the general case.
I've done some thinking about it and for coprime integers it's well-known you can divide coprime product by each of the factors: $2 \times 3 \times 5$ obviously divides numbers of the form $n(n+1)(n+2)(n+3)(n+4)$ because there's always a number in there divisible by $2$, one divisible by $3$, and one divisible by $5$. Not so obvious for four. You notice that either $n+2$ and $n+4$ or $n+1$ and $n+3$ have both a power of $2$ and a power of $4$ in them. So all in all you have tha $2 \times 3 \times 4 \times 5$ divides $n(n+1)(n+2)(n+3)(n+4)$. You can assemble special cases piecemeal with relative ease, but I can't see how it is at all clear that it is always the case that $n!$ will always divide $n$ consecutive integers.
We take it for granted that a sequence of $n$ consecutive integers always contains at least one which is divisible by each number up to $n$.
I just can't quite get around whether I can justify then jumping to the conclusion that $n!$ divides $n(n+1)(n+2)...(n+(n-1))$.