Power of n equalling (n - 1)! + 1

Theorem
There is exactly one solution to the equation in the integers:
 * $\paren {n - 1}! + 1 = n^k$

for $k > 1$, and that is:
 * $n = 5$
 * $k = 2$

Theorem
We have that:

Let $n > 5$.

Suppose $n$ is composite.

By Divisibility of n-1 Factorial by Composite n:
 * $n \divides \paren {n - 1}!$

and thus:
 * $n \nmid \paren {n - 1}! + 1$

showing that $\paren {n - 1}! + 1$ is not a power of $n$.

Suppose $n$ is prime.

Further suppose:
 * $\exists k \in \N: \paren {n - 1}! + 1 = n^k$

Then:

By Divisibility of n-1 Factorial by Composite n, since $n - 1 \ne 4$ and is composite:
 * $n - 1 \divides \paren {n - 2}!$

thus $k$ must be a multiple of $n - 1$.

However:

which shows that no multiple of $n - 1$ can satisfy our equation.

Hence there is no solution for $n > 5$.