Prove prime $p$ divides $\binom{p^k}{n}$ for any $n$, $1 \le n \le p^k-1$, $k$ a positive integer

Question

I have got $$\binom{p^k}{n} = \cfrac{p^k}{n} \binom{p^k-1}{n-1}=p\biggl[\cfrac{p^{k-1}}{n}\binom{p^k-1}{n-1}\biggl]$$ The first approach is to show that the term in the square bracket is also an integer. I know that $\binom{p^k-1}{n-1}$ is an integer, but how should I deal with $\frac{p^{k-1}}{n}$?

The second approach is from $$\binom{p^k}{n} = \cfrac{p^k}{n} \binom{p^k-1}{n-1}$$ we get$$n\binom{p^k}{n} = p^k\binom{p^k-1}{n-1}$$ Thus$$p^k \biggl| n\binom{p^k}{n}$$ Then we divide into two cases:
1) Suppose $gcd(p^k, n) = 1$, then by Euclid's lemma, we know $p^k\bigl|\binom{p^k}{n}$ and thus $p\bigl|\binom{p^k}{n}$.
2) Suppose $gcd(p^k, n) > 1$, then $p^k$ and $n$ have common divisors. Let $m$ be the product of all common divisors of $p^k$ and $n$, then $gcd(\frac{p^k}{m}, \frac{n}{m}) = 1$. From $\frac{p^k}{m} \bigl| \frac{n}{m}\binom{p^k}{n}$, we have $\frac{p^k}{m}\bigl|\binom{p^k}{n}$ where $\frac{p^k}{m}$ is an integer. How can I proceed from here?

Or is there any other approach? Thanks.

Answer 1

You have $n \binom{p^k}{n} = p^k \binom{p^k - 1}{n-1}$ so $p^k$ divides $n \binom{p^k}{n}$.

First case: $ \textrm{gcd}(n, p) = 1$. So we also have $ \textrm{gcd}(n, p^k) = 1$. According to Gauss theorem, we have $p^k | \binom{p^k}{n}$ and then $p | \binom{p^k}{n}$.

Second case: $ \textrm{gcd}(n, p) \neq 1$. We write $p = qs$ and $n = ms$ with $ \textrm{gcd}(m, q) = 1$ and $\textrm{gcd}(n, p) = s \geq 2$. $n \binom{p^k}{n} = p^k \binom{p^k - 1}{n-1}$ becomes $m s \binom{p^k}{n} = q^k s^k \binom{p^k - 1}{n-1}$ then $m \binom{p^k}{n} = q^k s^{k-1} \binom{p^k - 1}{n-1}$ so we get $q | \binom{p^k}{n}$. I do not know how to go further...

Answer 2

If $p$ is even and $k=1$, you have $\binom{p}{2} = \dfrac{p(p-1)}{2} \in \mathbb{N}$. If $\dfrac{p(p-1)}{2} = l p$ for some $l$, we get $2l = p-1$ thus $p$ is odd. This is absurd. So, necessarily you have to take either $p$ odd or $k \neq 1$.