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@ColinTheMathmo

Cancel the (n-k)!, and we're left with "why does k! divide the falling factorial \(n^{\underline{k}}\)?" (notation aside). And that can be answered, I think, with an understanding of divisibility by prime powers in ranges of consecutive integers.

Trickier, sure, but I think my vague approach above works. Lemme try to describe some more (maybe a proofinatoot later); it's similar to the classic exercise that shows that the power of prime \(p\) in \(k!\) is \(\sum_{i\geq1} \left\lfloor\frac k{p^i}\right\rfloor\) \) (which I now realize has a name, "Legendre's formula"). The idea extends to say that the power of prime \(p\) in the falling factorial \(n^{\underline{k}}\) is _at least_ that quantity.

To be clear, I agree with the other thread here, that the other definitions are more natural. I'm approaching this from "how can we show this directly?".

@bmreiniger I agree that this approach can be pushed through, but it's not immediately obvious, and does require care. The question then is: "How clear *can* this be made?"

The discussion on Twitter has been interesting:

ColinTheMathmo@ColinTheMathmo@mathstodon.xyz@bmreiniger You can show that each factor in the denominator divides the numerator, but showing that they all do so without fatally interfering with each other is a little trickier.