A simple binomial identity

$$\binom{p^n}{k} = \frac{p^n (p^n -1) ... (p^n -k+1)}{k!}= \frac{p^n}{k} \cdot \frac{ (p^n -1) ... (p^n -k+1) }{(k-1)!} = \frac{p^n}{k} \cdot \underbrace{\binom{p^n -1}{k-1}}_{\in \mathbb{Z}} $$

$p \Big| p^n$ and because $k<p^n$ we get that $$p \Big| \binom{p^n}{k}$$ Q.E.D

In this answer is is shown that the number of factors of $p$ that divide $\binom{n}{k}$ is $$ \frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}\tag{1} $$ where $\sigma_p(n)$ is the sum of the digits in the base-$p$ representation of $n$.

Since $\sigma_p\!\left(p^n\right)=1$ and $k,p^n-k\ne0$, we have that $\sigma_p(k),\sigma_p\!\left(p^n-k\right)\ge1$. Thus, $$ \frac{\sigma(k)+\sigma\!\left(p^n-k\right)-\sigma\!\left(p^n\right)}{p-1}\gt0\tag{2} $$ Therefore, the number of factors of $p$ that divide $\binom{p^n}{k}$ is greater than $0$.

Just a quick remark after the fact: If you accept that $$ (a +b )^{p} \equiv a^p +b^p\pmod p ,$$ for $a$ and $b$ indeterminants, then $$(a+b)^{p^n} = \left(\ (a + b )^p\ \right)^{p^{n-1}}\equiv \left(\ a^p + b^p\ \right)^{p^{n-1}}\equiv a^{p^n}+ b^{p^n}\pmod p,$$ which also gives the result.