Is this a valid proof? Modular arithmetic with fractions

Question

The problem is: Prove that for each prime $p$ and positive integer $n$, that $p^n$ divides $\ \binom{p^n}{p} - p^{n-1}$.

We have that

$\binom{p^n}{p} - p^{n-1} = p^{n-1}\left( \frac{(p^{n}-1)...(p^{n}-p+1)}{(p-1)...2 \cdot 1} - 1\right)$

Due to periodicity, each factor in the denominator will divide factors in the numerator in a one-to-one correspondence, so the fraction is an integer. It results to show that

$\frac{(p^{n}-1)...(p^{n}-p+1)}{(p-1)...2 \cdot 1} - 1 \equiv 0 \bmod p$

Now comes the part im not confident about. Wilson's theorem says that

$(p-1)! \equiv -1 \bmod p$

So

$\frac{(p^{n}-1)...(p^{n}-p+1)}{(p-1)...2 \cdot 1} - 1 \equiv \frac{(p-1)!}{(p-1)!} - 1 \equiv 1 - 1 \equiv 0 \bmod p$

But I'm not completely convinced when it comes to dealing with fractions modulo $p$. Is it for example valid to say that it follows from Wilson's theorem that

$\frac{1}{1\cdot...\cdot(p-1)} \equiv -1 \bmod p$

? It doesnt really make sense to me.

Answer 1

Since $p^n\equiv p\pmod p$, you have that $$(p^{n}-1)(p^{n}-2)\cdots (p^{n}-(p-1))\equiv (p-1)(p-2)\cdots 1\pmod p$$

So $$\frac{(p^{n}-1)(p^{n}-2)\cdots (p^{n}-(p-1))}{ (p-1)(p-2)\cdots 1}\equiv 1\pmod p$$

(Which is only true since $(p-1)!\neq 0\pmod p$.)

Answer 2

It is possible to say that $$(p-1)! \equiv -1 \pmod p \iff \frac 1 {(p-1)!} \equiv -1 \pmod p$$ since -1 is a modular multiplicative inverse of -1, and since $p$ is prime then it is the unique modular multiplicative inverse.