To prove that $(n-1)!+1$ is not a power of $n$.

Your solution is correct. Without Wilson's theorem, maybe you'd like to explain why $n\mid (n-1)!$ for all composite all $n$.

$\bullet$ If $n=p^2$, for some prime $p\ge 3$, then $$ n=p^2\mid p\cdot (2p)\mid (p^2-1)!=(n-1)!. $$ $\bullet$ If $n$ has at least two prime factors or $n=p^k$ for some prime $p$ and integer $k\ge 3$, then $n=ab$, for some $1<a<b<n$, hence $$ n=ab \mid (ab-1)!=(n-1)!. $$

Using Wilson's theorem, we see that if $n$ is composite, then $(n-1)!\not\equiv -1\mod{n}$, so that in fact $(n-1)!+1$ is not even divisible by $n$, much less a power of $n$.