Proof of Wilson's theorem via finite fields?
By Vieta's Formulas we know that the constant term of a polynomial is equal to the product of the roots of the polynomial when the degree is even, and the negative of that when the degree is odd. Since $\mathbb{F}_p^\times$ is cyclic of order $p-1$, any non-zero element of $\mathbb{F}_p$ satisfies $x^{p-1}-1=0$. Handling the $p=2$ case separately, we see that $p$ is odd so $p-1$ is even, so $-1$ is the product of all the roots of $x^{p-1}-1$. We have already observed $p-1$ unique roots to this equation, and since $\mathbb{F}_p$ is a field, it has exactly $p-1$ roots and so the roots are precisely the the non-zero elements with no repeats and we are done.
In the comments, it is mentioned that this holds not just in $\mathbb{F}_p$ but generally in $\mathbb{F}_q$. This proof generalizes because, by construction, the elements of $\mathbb{F}_q$ satisfy $x^q-x=0$. Dividing both sides by $x$ and continuing as stated in the above proof yields the result. In the general theorem, the $p=2$ case still needs to be handled separately and is less trivial.
NB: This is the same concept as your proof 2, but justified using language of Field Theory. In general, the boundary between "Galois Theory Concepts" and "number theory concepts" is pretty loose when it comes to Modular Arithmetic.
My guess is that you may be expected to use cyclicity of $\Bbb{F}_p^*$. Let $g$ be a generator of this cyclic group of order $p-1$. Then there are two ways of listing the elements of the multiplicative group (in different orders): $$ \Bbb{F}_p^*=\{1,2,3,\ldots,p-1\}=\{1,g,g^2,\cdots,g^{p-2}\}. $$ This implies that in $\Bbb{F}_p$ we have $$ \begin{aligned} (p-1)!&=1\cdot2\cdot3\cdots(p-1)\\ &=1\cdot g\cdot g^2\cdots g^{p-2}\\ &=g^{0+1+2+3+\cdots+(p-2)}. \end{aligned} $$ In the last form the exponent is $$ E=\sum_{j=0}^{p-2}j=\frac12(p-2)(p-1). $$ Because $p$ is odd, we see that $E$ is an odd multiple of $(p-1)/2$.
On the other hand, by cyclicity, $g^{(p-1)/2}$ is of order two. As $-1$ is the only element of order two, we thus have $$ W:=g^{(p-1)/2}=-1, $$ and $$ g^E=W^{p-2}=(-1)^{p-2}=-1. $$