Combinations of sets raised to the power of a prime modulus: $\binom{p^{\alpha}-1}k \equiv (-1)^k \pmod p$

Question

This is a problem out of the text Introduction to the Theory of Numbers by Niven, Zuckerman, and Montogmery and I am having quite a bit of trouble with it. I tried to prove it directly, but that didn't work. The question is:

Let $p$ be a prime and $\alpha$ a positive integer. Show that $\binom{p^{\alpha}-1}k \equiv (-1)^k \pmod p$ for $ 0 \leq k \leq p^\alpha -1$.

If relevant, I did prove that ${{p^{\alpha}}\choose k} \equiv 0 \pmod p$ for $ 0 < k < p^\alpha $ but I can't seem to incorporate it.

Answer 1

$${p^a-1\choose k}={(p^a-1)(p^a-2)\cdot\dots\cdot(p^a-k)\over(1)(2)\cdot\dots\cdot(k)}\equiv{(-1)(-2)\cdot\dots\cdot(-k)\over(1)(2)\cdot\dots\cdot(k)}=(-1)^k$$ where the congruence is modulo $p$.