If $p$ is prime then $p^{n-1}\mid \binom{p^n}{p}$.

Question

I am stuck with this problem:

If $p$ is prime then $p^{n-1}\mid \binom{p^n}{p}$.

The thing is that I don't know much properties of binomial coefficients and I'd accepts hints.

Answer 1

By Kummer, the $p$-adic valuation of $\binom{p^n}{p}$ is equal to the number of "carry-overs" when adding $p^n-p$ and $p$ in base $p$. The base $p$ expression of $p^n-p$ is $$0\cdot p^n +(p-1)\cdot p^{n-1}+\cdots+(p-1)\cdot p^2+(p-1)\cdot p+0\cdot p^0$$ so adding $p^n-p$ and $p$ in base $p$ yields $n-1$ carry overs (one from the $p^1$ term, then this gives a carry over from the $p^2$ term, and so on until we get a carry over from the $p^{n-1}$ term).

Answer 2

Hint:

The binomial coefficient $\;\dbinom ak$ is the number of arrangements $\;A_n^k=n(n-1)(n-2)\dotsm(n-k+1)$, divided by $k!$. So here, you have

$$\binom{p^n}p=\frac{p^n(p^n-1)(p^n-2)\dotsm(p^n-p+1)}{p!}=\frac{p^{n-1}(p^n-1)(p^n-2)\dotsm(p^n-p+1)}{(p-1)!}$$ Can you see why $(p-1)!$ is coprime with $p^{n-1}$?

Answer 3

The binomial coefficient can be defined as $$\binom{p^n}{p}:=\frac{p^n\cdot (p^n-1)\cdot \ldots\cdot(p^n-p+1)}{p!}$$ Observe finally that

\begin{align*} \frac{p^n\cdot (p^n-1)\cdot \ldots\cdot(p^n-p+1)}{p!\cdot p^{n-1}}&=\frac{(p^n-1)\cdot \ldots\cdot (p^n-p+1)}{(p-1)!}\tag{1}\\ &\equiv \frac{(p^n-1)\cdot \ldots\cdot (p^n-p+1)}{-1}\pmod p\tag{2}\\ &\equiv (p^n-1)\cdot \ldots\cdot (p^n-p+1)\pmod p \end{align*}

The transition from $(1)$ to $(2)$ is a consequence of Wilson's Theorem.