Does $((x-1)! \bmod x) - (x-1) \equiv 0\implies \text{isPrime}(x)$ Does $$((x-1)! \bmod x) - (x-1) = 0$$
imply that $x$ is prime?

I want to add that this is the easy direction of Wilson's theorem.  I have often seen Wilson's Theorem stated without the "if and only if" because this second direction is so easy to prove:
Proof: If $x > 1, x \ne 4$ were not prime, then the product $(x-1)!$ would contain two factors multiplying to get $x$, and thus we would have $(x - 1)! \equiv 0 \pmod x$, contradicting the statement.  On the other hand if $x = 4$, then $(x - 1)! = 6 \equiv 2 \not \equiv -1$ so in any case $x$ must be prime.
At one point $4$ was my favorite integer for precisely the reason that it was the only exception to $(x - 1)! \equiv 0 \text{ or } 1 \pmod x$.
