Post History

You started on the right track with $n(n - 1)\ldots(n - [k - 3])(n - [k - 2])(n - [k - 1])$.

What you should have noticed is that there are $k$ terms in that product. The first term subtracts 0 from $n$, the second term subtracts 1, and so on, with the $i$th term subtracting $i - 1$, all the way up to the $k$th term subtracting $k - 1$.

So the logical extension of this pattern to $k = 1$ is a product with only one term, the initial one, which subtracts 0 from $n$... which is just $n$.