A simple binomial identity Is there a simple way of showing that a prime $p$ must divide the binomial coefficient $p^n\choose{k}$ for all $n\geq 1$ and $1\leq k\leq p^n-1$?

$$\begin{align*}\binom{p^n}{k} = \frac{(p^n)!}{k! (n-k)!} &= \frac{(p^n)(p^n-1)(p^n-2)(p^n-3)\cdots (p^n-(k-1))}{1\cdot 2\cdot 3 \cdots (k-1)k} \\
&= \frac{p^n}{k}\cdot \frac{p^n-1}{1} \cdot \frac{p^n-2}{2}\cdots \frac{p^n-(k-1)}{k-1} \\
&= \frac{p^n}{k} \cdot \prod_{i=1}^{k-1} \frac{p^n-i}{i}.\end{align*}$$
(Note that these smaller fractions are not necessarily integers.)  
We are trying to show that after all cancellations of prime factors, there remains at least one factor of $p$ in the numerator of this large fraction.  
Consider the fractions of the form $\frac{p^n-i}{i}$.
Now $p^n-i$ is divisible by $p^j$ if and only if $i$ is divisible by $p^j$, for all $1\leq j\leq n$.  So in each fraction $\frac{p^n-i}{i}$, all the $p$'s in the numerator and denominator cancel perfectly.
So in the entire fraction, the only remaining $p$'s in the numerator are those in $p^n$, and the only remaining $p$'s in the denominator (if any) are factors of $k$.  But $k<p^n$, so $k$ can have at most $n-1$ factors of $p$, leaving at least one left over in the numerator.
