${p^k \choose p^{k-1}} \neq 0 \pmod {p^k}$ 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.
 A: For a binomial coefficient $\binom{m}{n}$ where $1 \leq n \leq m$, show $\binom{m}{n} = \frac{m}{n}\binom{m-1}{n-1}$.  Therefore $\binom{p^k}{p^{k-1}} = p\binom{p^k-1}{p^{k-1}-1}$.  Use Kummer's theorem (mentioned in a comment above) to show $\binom{p^k-1}{p^{k-1}-1} \not\equiv 0 \bmod p$ by writing $p^k-1$ and $p^{k-1}-1$ in base $p$.
EDIT:  When I wrote "Kummer's theorem" I actually had in mind Lucas' theorem about binomial coefficients, which expresses a binomial coefficient mod $p$ in terms of binomial coefficients of base $p$ digits. if $m  = m_0 + m_1p + \cdots + m_dp^d$ and $n = n_0 + n_1p + \cdots + n_dp^d$ where $0 \leq m_i, n_i \leq p-1$ then $\binom{m}{n} \equiv \binom{m_0}{n_0}\binom{m_1}{n_1} \cdots \binom{m_d}{n_d} \bmod p$. Apply this with $m = p^k - 1$, $n = p^{k-1}-1$, and $d = k-1$.
A: This was already mentioned in the comments, but it's simple enough I figured I'd type it up. Since you can see Kummer's theorem as coming from Legendre's formula, $v_p(n!) = \frac{n-s_p(n)}{p-1}$ where $v_p$ is the p-adic valuation and $s_p$ is the sum of digits in base $p$, I'll just go from there.
First split it up by additivity of the valuation,
$$v_p\left(\binom{p^k}{p^{k-1}}\right) = v_p((p^k)!)-v_p((p^{k-1})!) - v_p((p^k-p^{k-1})!)$$
Now use Legendre's formula to simplify
$$=\frac{s_p(p^{k-1})+s_p((p-1)p^{k-1})-s_p(p^k)}{p-1}$$
Since these numbers are in base $p$, they each have a single nonzero digit and simplify nicely to,
$$=\frac{1+(p-1)-1}{p-1} = 1$$
So actually we have proven the stronger result that for $k\ge 2$,
$$\binom{p^k}{p^{k-1}} \not \equiv 0 \mod p^2$$
A: I think you should go  with
$$v_{p}(\binom{p^{k}}{p^{k-1}})=v_{p}(p^{k}!)-v_{p}(p^{k-1}!)-v_{p}((p^k-p^{k-1})!)=\sum_{i=1}^{k}p^{k-i}-\sum_{i=1}^{k-1}p^{k-1-i}-\sum_{i=1}^{k-1}\left[p^{k-i}-p^{k-1-i}\right]=1$$
