We have $(n-1)! = n^{k-1} - 1$ so $(n-2)! = n^{k-2} + \ldots + n + 1$. Now for $n\geq7$ the factors $2$ and $\left\lfloor \frac{n-1}{2}\right\rfloor$ are distinct and do not exceed $n-2$, so LHS is divisible by $n-1$ while each term in RHS gives remainder $1$. Thus $n-1$ divides $k-1$ and in particular $k\geq n$ but $2n! > n! + n = n^k \geq n^n$ is false for $n\geq 7$. If $n=1$, no solutions. If $n=2$ or $n=3$, then $k=2$. For $n=4,6$ no solutions while $n=5$ works with $k=3$.

Also here: https://artofproblemsolving.com/community/c6h3065883p27661925