I would like to show ${p^k \choose p^{k-1}} \neq 0 \pmod{p^k}$, where $p$ is a prime and $k >1$. I think this is a rather obvious results and I cannot seem to prove it unfortunately. I have looked at some numerical data. Usually, for the first values of $k$, ${p^k \choose p^{k-1}} \equiv p \pmod{p^k}$, but this pattern does not hold for all values; however, it is still true that they are not congruent to $0$. This is what I have done:
$${p^k \choose p^{k-1}} = \frac{p^k...(p^{k-1})!}{(p^{k-1})!\cdot (p^{k} - p^{k-1})!}= \frac{p^k...(p^{k-1}+1)}{(p^{k-1}(p-1))!}$$
I am not sure how to proceed form here to show $p^k$ does not divide this.
Update:
I am not familiar with Kummer's Theorem or Legendre's formula. While it is not extremely difficult to learn about them, I would prefer a solution that does not rely on these concepts.