Let . Show that is a multiple of for .
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