Wilson's theorem Prime Generator

Question

There is a famous theorem in number theory called Wilson's Theorem.

Statement: $n$ satisfies, $(n-1)! + 1 = 0\pmod n$ if and only if $n$ is prime.

Another way of looking at the statement is that, $\dfrac{(n-1)! + 1}{n}$ is an integer only if $n$ is prime.

I have noticed a pattern in these integers. They always come out to be prime. I have made a code, to test primes upto a large number, and all of the integers that have come from the fraction, is found to be prime. If the aforementioned statement always holds true for all primes, then it would behave as a prime generator. I haven't found any useful information currently existing on the internet.

Hence, I am looking for a mathematical proof for the following:

Let $f:\mathbb{N}\to\mathbb{N}$ be defined as, $$f(n) = \frac{(n-1)! + 1}{n}$$for all $n\in\mathbb{N}$.

Prove that: $$f(n)\in \mathbb{P};\ \ \forall n \in\mathbb{P}$$ Where $\mathbb{P}$ is the set of Primes.

Now, I do know that for $n = {2,3},\ f(n) = 1$. This should be ignored as sometimes, $1$ trivially shows up while working with primes.

Answer 1

It is not always prime: $ f(13) = 13 \times 2834329 $. For the primes $ 2 \leq p \leq 997 $, the only time $ f(p) $ is (probably) prime is for $ p = 5, 7, 11, 29, 773 $.

Answer 2

That's false. First counterexample $13$.

$$\frac{(13-1)! +1}{13} = 36846277 = 13 \times 2834329$$

And if you do not like that it is $13$ that shows up again take $17$.

$$\frac{(17-1)! +1}{17} = 1230752346353 = 61\times 137\times 139 \times 1059511$$

Or also $19$ and $23$. Then for $29$ it is prime again.

A reason why it is plausible that it is true for small $n$ frequently is that $(n-1)!+1$ cannot be divisible by any primes less than $n$. But otherwise there is not much reason for this to be prime, though it should still be prime a bit more frequently than a typical number of that size.