The (little) Fermat Theorem:
if $p$ is prime then $a^{p-1}=1\ \text{mod} \ p$ for whatever $a$
may be easily understood as a consequence of the group structure of multiplication on integers mod $p$.
Considering addition and multiplication, the stucture is extended to a ring, where Lagrange's lemma on the identity of polynomials with identical roots holds. Wilson's identity can then be seen as a consequence of the ring structure: since both polynomials of degree $p-1$
$P(z)=(z-1)(z-2)...(z-(p-2))(z-(p-1))$ and
$Q(z)=z^{p-1}-1$
have the same roots mod $p$ (the first by construction, the second by virtue of Fermat's Theorem) it follows that their free terms should be equal:
if $p$ is prime then $(p-1)!=-1\ \text{mod} \ p$
which would almost be Wilson's theorem. This last is in fact stronger since it includes the reciprocal too, being therefore a primality test for $p$.
My actual question is: when writing the above demonstration, why are not usually equated the remaining coefficients of $z^k$ in order to obtain a host of Wilson-like identitites? Considering the coefficient of $z$ would thus obtain
$\sum_{j=1}^{p-1}({1\cdot 2\cdot 3 ...\not{j}...(p-2)\cdot(p-1)})=0\ \text{mod}\ p$
or equivalently
$\sum_{j=1}^{p-1}\ \frac{1}{j}=0\ \text{mod}\ p$.
The general form derived from the coefficient of $z^k$ would then read
$H_k(p-1)=\sum_{j=1}^{p-1}\ \frac{1}{j^k}=0\ \text{mod}\ p$, which holds for $p > k$.
Here $H_k(p-1)$ is the harmonic number of order $k$, which accordingly has to be 0 mod $p$ if $p$ is prime.
How are these identities called and why are they not mentioned in the context of Wilson' Theorem?