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.

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.
