Unable to find logic behind question on primes. Let $p$ be a prime number and $s$ be a positive integer. Show that for any $i \in \{0, 1, . . . , p^s− 1\},\ \binom{p^s−1}i \equiv (−1)^i(mod\ p)$.
Formula-wise the question attempts to take the ratio of:
$\frac{(p^s−1)!}{i!(p^s−1-i)!}$, for all $i\in \{0, 1, . . . , p^s− 1\}$.
On taking some special values of $i$, get the below:
1. If $i=0$, the ratio =$1$, 
2. If $i=1$, the ratio =$p^s−1$, 
3. If $i=p^s−1$, the ratio =$1$, 
But am unable to understand what is the logical implication of the question; and why only prime value of $p$ is considered.
Also, suggest an approach to solve the same.
 A: It suffices to show that $\binom{p^s-1}{i+1}\equiv(-1)\binom{p^s-1}i\pmod p$.
Write $i+1=p^kr$ where $p\not\mid r$. Since $p$ is prime and $p\not\mid r$, we have $\gcd(p,r)=1$, so there exists an integer $r^{-1}$ such that $r^{-1}r\equiv1\pmod p$. Also note that $k<s$, so $p^{s-k}\equiv0\pmod p$.
Without reducing mod $p$, we can verify the following equality of integers:
$$
r\binom{p^s-1}{i+1}=(p^{s-k}-r)\binom{p^s-1}i
$$
Then multiplying both sides by $r^{-1}$ and reducing mod $p$ using the observations above gives the desired result.
A: I completely agree with Siong Thye Goh's answer and consider his answer more elegant than the one I am about to present.  My answer is intended to show the much more pedestrian and inelegant but still viable approach of induction.
Assume that the conjecture is true for $i$, where 
$0 \leq i \leq p^s - 2.$
Then $\;\binom{p^s - 1}{i + 1},\;$ which is an integer, equals
$\;\binom{p^s - 1}{i} \times \frac{p^s - [i+1]}{i + 1}.$
By inductive assumption, $\;\binom{p^s - 1}{i} \;\equiv (-1)^i \pmod{p}.$
Further, $\;(-1)^i [-(i+1)] \equiv (i+1)(-1)^{(i+1)} \pmod{p}
\;\Rightarrow$
$\;\binom{p^s - 1}{i} \times [p^s - (i+1)]
\;\equiv\; (i + 1) (-1)^{(i+1)}\pmod{p} \;\Rightarrow$
$\;\binom{p^s - 1}{i} \times \frac{p^s - [i+1]}{i + 1}
\;\equiv\; (-1)^{(i+1)}\pmod{p}.$
$\underline{\text{Addendum}}$
As Karl indicated in his comment below, I must examine the case of 
$\;(i+1) \;\equiv\; 0 \pmod{p}\;$ separately.  I am working on it and will update my answer if and when I find a remedy.
Convoluted remedy found : proof by contradiction.
Let $pk$ denote the smallest multiple of $p$ such that the conjecture is false.  
Then, from the induction used in the first part of this answer, the conjecture is true for $i = (pk - 1).$
Let $v_p(n) \;\equiv\;$ the largest exponent $\theta$ such that 
$p^{\theta} | n.$ 
Let $v_p[(p^s - 1)!] = \alpha, \;v_p[(pk - 1)!] = \beta, 
\;v_p[(p^s - pk)!] = \gamma,\;$ and $\;v_p(pk) = \delta.$ 
By assumption, $\;\binom{p^s - 1}{pk-1} \;\equiv\; (-1)^{(pk-1)}
\;\pmod{p}.$ 
Therefore, $\;\alpha = \beta + \gamma,\; 
(p^s - 1)!\;$ has form $\;(p^\alpha) \times a,$ 
$(pk-1)!(p^s - pk)!\;$ has form $\;(p^\alpha) \times b,$ 
where $p$ does not divide $a$ or $b$ and
$a \equiv b\times (-1)^{(pk-1)} \pmod{p}.$
$v_p(pk) = \delta \;\Rightarrow v_p(p^s - pk) = \delta.$ 
Further, $\;pk = p^\delta \times r,\;$ where 
$p$ does not divide $r \;\Rightarrow\;$ 
$(p^s - pk)\;$ will have form $(p^\delta) \times (-t),\;$
where $\;t \;\equiv\; r \pmod{p}.$
Thus, $\;(pk)!(p^s - 1 - pk)! \;=\; $
$(pk) \times [(pk-1)!]\times\frac{(p^s - pk)!}{(p^s - pk)},$ 
which can be re-expressed as $(p^\delta \times r) \times (p^\alpha \times b)$
$\frac{1}{p^\delta\times(-t)} \;\Rightarrow\;$ 
$\binom{p^s - 1}{pk} \;\equiv\; (-1)^{(pk-1)} \times \frac{-t}{r} 
\;\equiv\; (-1)^{(pk)} \pmod{p}.$ 
This yields a contradiction. Therefore, the conjecture also holds for any $i < p^s - 1,$ 
where $i$ is a multiple of $p$.
A: Let $$v=\binom{p^s-1}{i} = \frac{\prod_{j=1}^{i}(p^s-j)}{i!}$$
Suppose $j \in \{1, i\}$, we can write $j=p^{r_j}w_j$ where $gcd(p,w_j)=1$, $r_j < s$, then $$p^s-j=p^s-p^{r_j}w_j=p^{r_j}(p^{s-r_j}-w_j)$$
$$v=\prod_{j=1}^i\left(\frac{p^s-j}{j}\right)=\prod_{j=1}^i\left( \frac{p^{r_j}(p^{s-r_j}-w_j)}{p^{r_j}w_j}\right)=\prod_{j=1}^i\left( \frac{p^{s-r_j}-w_j}{w_j}\right)$$
$$v\cdot \prod_{j=1}^iw_j = \prod_{j=1}^i (p^{s-r_j}-w_j)$$
$$v\cdot \prod_{j=1}^iw_j \equiv  \prod_{j=1}^i (p^{s-r_j}-w_j) \equiv  \prod_{j=1}^i (-w_j)\equiv (-1)^i \prod_{j=1}^iw_j \pmod{p}$$
Since $gcd(p,w_j)=1$, $\left(\prod_{j=1}^iw_j\right)^{-1}$ exists, multiplying it on the both sides, we have 
$$v \equiv (-1)^i \pmod{p}$$
That is 
$$\binom{p^s-1}{i}\equiv (-1)^i \pmod{p}$$

Remark: I made a mistake earlier for not checking if an element is invertible.
This is the wrong approach:
\begin{align}
\binom{p^s-1}{i} &= \left(\prod_{j=0}^{i-1}(p^s-1-j)\right)(i!)^{-1}\\
&\equiv \left(\prod_{j=0}^{i-1}(-1-j)\right)(i!)^{-1} \equiv (-1)^i(i!)(i!)^{-1}  \\&\equiv (-1)^i \pmod{p}
\end{align}
because I did not check that $(i!)^{-1}$ exists and indeed it need not exists.
In my latest approach, I ensure that $gcd(w_j,p)=1$ and hence $(\prod_j w_j)^{-1}$ exists. Also, by expressing $j=p^{r_j}w_j$, I have illustated that the $p^{r_j}$ factor  in $p^s-j$ can cancel out with $p^{r_j}$ factor in $j$. 
That is now I use the property that the 


*

*$p^{r_1}$ in $1$ and $p^s-1$ cancels out.

*$p^{r_2}$ in $2$ and $p^s-2$ cancels out


and so on.
