Show that $\binom{p^a-1}{k}\equiv(-1)^k\pmod{p}$ for $0\le k\le p^{a}-1$.
Solution: $\binom{p^a-1}{k}=\frac{(p^a-1)(p^a-2)\cdots(p^a-k)}{k!}$.
Idea: Then power of $p$ in $k!$ must be less than the power of $p$ in Numerator. ----(**)
Then i can directly write $p^a-1\equiv-1\pmod{p}$,$p^a-2\equiv-2\pmod{p}\cdots$ and $p^a-k\equiv-k\pmod{p}$.
Then, we have: $\binom{p^a-1}{k}\equiv(-1)^k\pmod{p}$.
Is my idea correct? If yes, how should I proceed to prove (**) statement.
Thanks in advance.