Euler is welcomed again

Here one may appeal to the generalized Wilson's theorem which states that the product of the units mod $n$ is equal to $-1$ if and only if there exists a primitive root mod $n$ . If this is the case then $\phi^! (n) + 1 \equiv -1 + 1 \equiv 0 \mod n$ as desired. Then, to find these numbers, one need only apply the criterion that there exist primitive roots mod $n$ if and only if $n=1,2,4,p^k, 2p^k$ where $p$ is an odd prime. Then for a solution, all one needs is a list of primes smaller than 50 and then to find all the numbers smaller or equal to 50 that are expressible in the form just presented.

Bonus: For fun, if someone wanted a way to compute $\phi^! (n)$ without needing to list all the integers relatively prime to $n$ then I have proven an alternative that only needs one to list all the square-free divisors of $n$ :

$\log(\phi^!(x)) = \log \left( \prod_{n \leq x, gcd(n,x)=1} n \right) = \sum_{n \leq x, gcd(n,x)=1} \log(n) = \sum_{d | x} \mu (d) \sum_{q \leq \frac{x}{d}} \log(qd) = \sum_{d | x} \mu (d) \log \left( \prod_{q \leq \frac{x}{d}} qd \right) = \sum_{d | x} \mu (d) \log \left( \left(\frac{x}{d} \right)! d^{\frac{x}{d}} \right) = \\ \sum_{d | x} \log \left(\left( \left(\frac{x}{d} \right)! d^{\frac{x}{d}} \right)^{\mu (d) }\right)$ and therefore $\phi^! (x) = \prod_{d | x} \left( \left[ d^{\frac{x}{d}} \left( \frac{x}{d} \right)! \right]^{\mu(d)} \right)$