Proof $k \equiv (−1)^{n} \pmod{p}$ [duplicate]

I have $k$ defined as: $$ k \equiv \frac{(np)!}{n!p^{n}} $$

where p is prime and k is positive, working ahead I've also found $ k = (-1)^{n}\pmod{p} $, I ideally need to trace back to actually prove this fact but from the first statement, any ideas on how to do so would be really appreciated.

Solution 1:

First I recognize that this product with factors of $p$ missing is similar to a product you'd use if you wanted to generalize Wilson's theorem,

$$f(n)=\prod_{\substack{1\le k \le n \\ p \nmid k}}k = \frac{\prod_{1\le k \le n}k}{\prod_{\substack{1\le k \le n \\ p \mid k}}k} = \frac{n!}{\prod_{k=1}^{\lfloor n/p\rfloor}pk}= \frac{n!}{\lfloor n/p\rfloor ! p^{\lfloor n/p\rfloor}} $$

You can ignore this next part as it's irrelevant to the actual derivation but, you should see that this generalizes Wilson's theorem with $f(p^n-1)\equiv -1 \mod p^n$ because it's the product of all the units in $\mathbb{Z}/p^n\mathbb{Z}$ and $-1$ is the only number that doesn't have its inverse to pair with. I only state this because it explains how I recognized it so it didn't look completely pulled from thin air.

Now let's continue onwards. Let's examine $f(pn)$.

$$f(pn)=\prod_{\substack{1\le k \le pn \\ p \nmid k}}k =\frac{(pn)!}{n! p^n} $$

Now we have the actual thing we want. If we look at the product modulo $p$, we can split it up into $n$ products that all go from $k=1$ to $k=p$, however we can exclude the $k=p$ term as $p \nmid k$,

$$\prod_{\substack{1\le k \le pn \\ p \nmid k}}k \equiv \prod_{i=0}^{n-1} \prod_{k=1}^{p-1}(k+pi)\equiv \left(\prod_{k=1}^{p-1}k\right)^n\mod p $$

Now we have $$(p-1)!^n \equiv (-1)^n \mod p$$