This problem is taken from Ivan Niven's "An Introduction to the Theory Of Numbers".
Show that ${{p^\alpha-1}\choose{k}} \equiv ({-1})^k\pmod p$.
Note: This is not similar to this one, as $k! | p^\alpha$ is possible.
My Attempt: We proceed by induction on k: Let, ${{p^\alpha-1}\choose{k-1}}=r$.Let $k=p^t*q \ : t<\alpha$. $(k,p^\alpha-k)=p^t$. $${{p^\alpha-1}\choose{k}} = {{p^\alpha-1}\choose{k-1}}*\frac{p^\alpha-k}{k}={{p^\alpha-1}\choose{k-1}}*\frac{p^{\alpha-t}-q}{q}$$.
So, by induction hypothesis, $r = q*(\frac{r}{q})\equiv(-1)^{k-1} \pmod p$. (As $(q,p^{\alpha-t})=1,\ \frac{r}{q} \in \mathbb Z$)So, $(p^{\alpha-t}-q)(\frac{r}{q})\equiv qq^{-1}(-1)^{k} \pmod p \equiv (-1)^k \pmod p$. Is the proof complete? Or did I miss something?
Alternative Proof of ${{p^\alpha-1}\choose{k}} \equiv ({-1})^k (mod \ p)$
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elementary-number-theory
proof-verification
modular-arithmetic
divisibility