AUT ABooklet

Answer. The only solutions are $(2, 2)$, $(3, 2)$ and $(5, 3)$.

Because of $n! + n > n$, we immediately get $k \ge 2$. We divide both sides of the equation by $n$ and get
$$(n - 1)! + 1 = n^{k-1}.$$
Now, we distinguish two cases:

* $n$ is not a prime.
Since $n$ is clearly not $1$, we can write $n$ as $n = ab$ for integers $a, b$ with $1 < a, b < n$ which implies $1 < a \le n - 1$ and therefore $a \mid (n - 1)!$. We conclude that $a > 1$ is relatively prime to the left-hand side $(n - 1)! + 1$, but $a$ divides the right-hand side $n^{k-1}$. This is not possible, so there are no solutions in this case.

* $n$ is a prime.
We check $n = 2, 3, 5$ and find the solutions $(2, 2)$, $(3, 2)$ and $(5, 3)$.
From now on, let $n \ge 7$. We get
$$\begin{align*} (n-1)! &= n^{k-1} - 1 \\ \implies (n-1)! &= (1 + n + n^2 + \dots + n^{k-2})(n-1) \\ \implies (n-2)! &= 1 + n + n^2 + \dots + n^{k-2} \end{align*}$$
Since $n$ is prime and bigger than $3$, the number $n-1$ is even and not a prime. Furthermore, $n-1$ is not the square of a prime since $4$ is the only even square of a prime and $n-1 \ge 6$. Therefore, we get $n-1 = ab$ with $1 < a, b \le n-1$ and $a \ne b$. We obtain that $(n-2)!$ contains the separate factors $a$ and $b$ and is therefore divisible by $ab = n-1$ which implies $(n-2)! \equiv 0 \mod (n-1)$. Furthermore, $n \equiv 1 \mod (n-1)$, and therefore
$$0 \equiv 1 + 1 + 1^2 + \dots + 1^{k-2} \equiv k - 1 \mod (n-1).$$
We conclude that $n-1$ divides $k-1$ and we write $k-1 = l(n-1)$ for a positive integer $l$. The case $k=1$ and $l=0$ has already been treated. Therefore, we get $k-1 \ge n-1$.
However,
$$(n-1)! = 1 \cdot 2 \cdot 3 \cdots (n-1) < \underbrace{(n-1) \cdot (n-1) \cdots (n-1)}_{n-1 \text{ times}} = (n-1)^{n-1},$$
and therefore
$$n^{k-1} = (n-1)! + 1 \le (n-1)^{n-1} < n^{n-1} \le n^{k-1},$$
giving a contradiction. So there are no further solutions.

(Michael Reitmeir) $\square$