1. ## Multiple of p

Let $r \in \mathbb{N}$. Show that $\binom{p^r}{k}$ is a multiple of $p$ for $0 < k < p^r$.

How to do this question..?

2. Originally Posted by usagi_killer
Let $r \in \mathbb{N}$. Show that $\binom{p^r}{k}$ is a multiple of $p$ for $0 < k < p^r$.

How to do this question..?

$\displaystyle \binom{p^r}{k}=\frac{\left(p^r-(k-1)\right)\left(p^r-(k-2)\right)\cdot \ldots \cdot p^r}{1\cdot 2\cdot \ldots \cdot k}$ , so the solution follows at once from the following

Lemma: For every $\displaystyle 1\le j < k$ , we have that $\displaystyle p^m\mid p^r-(k-j)\,\Longleftrightarrow\,p^m\mid k-j$ , for some $\displaystyle m\in\mathbb{N}$

From the above, it turns out that the powers of p cancel each others in pairs (one from the numerator, one from the denominator), and $\displaystyle p^r$ doesn't cancel COMPLETELY with no power of $\displaystyle p$ from the denominator EXCEPT, perhaps, with $\displaystyle k=ap^s$...as $\displaystyle k < p^r$ , it follows that $\displaystyle s<r$ and then $\displaystyle \frac{p^r}{k}=\frac{p^{r-s}}{a}$ and we're done.

Tonio

3. lol sorry I don't really get what you mean.

How did you prove they're multiples of p..?