A surprising constant

This problem is really asking for the product of the numbers in a row of Pascal's triangle modulo $p$ . For modulo $p$ , I will use the following pattern as demonstrated by modulo 5:

I'll call each group of ${ p }^{ k }$ rows ${ S }_{ k }$ . It is true that the ${ S }_{ k }$ 's that compose ${ S }_{ k+1 }$ are multiplied by their corresponding coefficients in the first $p$ rows. (Meaning every element of them is multiplied by that coefficient). Let $\displaystyle { P }_{ k } = \left(\prod _{ i=0 }^{ { p }^{ k }-1 }{ \binom{{ p }^{ k }-1}{i} } \quad ( mod \quad p) \right)$ Then we know that $\displaystyle { P }_{ k } = { { P }_{ k-1 } } ^ { p } { P }_{ 1 }\quad ( mod \quad p )$ This is due to the pattern mentioned above: we have the product of the last row of ${ S }_{ k-1 }$ $p$ times, and each of these is multiplied by a corresponding element in the ${p}^{th}$ row. ${ P }_{ k }$ is ${ P }_{ 1 }$ to some power in mod $p$ .

Luckily, ${ P }_{ 1 }$ is 1 mod $p$ if $p$ is $1$ mod $4$ . This can be shown with the factorial expansion: $\displaystyle { P }_{ 1 }=\prod _{ i=0 }^{ p-1 }{ \frac { (p-1)(p-2)...(p-i) }{ i! } }$ This shows that every element is $1$ or $p-1$ mod $p$ , and if $p$ is 1 mod 4 then the number of odd $i$ 's is even, so the entire expression is 1 mod $p$ . Thus, all ${ P }_{ k }$ are 1 mod $p$ with the given condition, so $K=1$ .

By Lucas's Theorem, we have that each of the binomial coefficients is congruent to $(-1)^i$ modulo p.But taking the product, we get that the exponent is an even number (using that p is 1 mod 4).So their product is 1 mod p.