The power

Observe that $n>1$ . Let $q$ be a prime divisor of $n$ . Then $q \mid n^m-(n-1)!=1$ . Contradiction. Thus $n$ is a prime number. Let $n=p$ be a prime number greater than $5$ . Then the equation becomes $(p-2)!=p^{m-1}+p^{m-2}+ \dots + p+1 \qquad (1)$ Notice that $2<\frac{p-1}{2}<p-2$ . Hence $p-1 \mid (p-2)!$ and from equation $(1)$ $p^{m-1}+p^{m-2}+ \dots + p+1 \equiv m \equiv 0 \pmod{p-1}$ Therefore, $p-1 \mid m$ and $p^m \ge p^{p-1}>(p-1)!+1$ . Contradiction. Thus $p \le 5$ .

Its easy to see that $(n, m)=(p, m)=(2, 1), (3, 1), (5, 2)$ are solutions.

Relevant wiki: Lifting the exponent lemma

The couples are $(2,1), (3,1), (5,2)$ .

Rewrite the equation as $(n-1)! = n^m-1$

Then let's start to consider every possible couple where $n>6$ . The main idea consists to analyze the factorization of the LHS and of the RHS of the equation, focusing on the power of a single prime factor.

In the left-hand side the power of a generical prime $p$ corresponds to: $\sum _{k=1} ^{ \infty} \lfloor \frac {n-1}{p^k} \rfloor$

In the right-hand side the power of a generical prime $p \ne 2$ corresponds to: $v_p (n^m-1)$ Thus, assuming that $(p,m) = 1$ , $v_p (n^m-1) = v_p (n^m-1^m) = v_p (n-1)$ Since the two sides of the equation must be equal, then the factorization must be equal too and then also the power of every prime factor. In particular: $\sum _{k=1} ^{ \infty} \lfloor \frac {n-1}{p^k} \rfloor = v_p (n-1)$ Define $\alpha = v_p (n-1)$ , so: $\sum _{k=1} ^{ \infty} \lfloor \frac {n-1}{p^k} \rfloor \ge 1+p+p^2+ \dots+p^{\alpha-1} > 1+2+2^2+ \dots +2^{\alpha-1} = 2^{\alpha}-1 > \alpha = v_p (n-1)$ $\rightarrow \sum_{k=1} ^{ \infty} \lfloor \frac {n-1}{p^k} \rfloor > v_p (n-1)$ If $\alpha >1$ for at least one prime of the factorization of $(n-1)!$ , then there aren't solutions: this happens when $3^2|(n-1)! \rightarrow n>6$ , since $3^2$ is first power of a prime with $\alpha>1$ and $p \ne 2$ .

Finally, we must analyze the case $(p,m) > 1$ for every prime until $n-1$ (otherwise we would go back to the previous case) . Let $m = \prod _{i=0} ^k p_i$ where $p_k$ is the greatest prime $\le n-1$ ; then: $(n-1)! < n! < n^n \le n^m-1$ since $n<\prod _{i=0} ^k p_i$ if $n \ge 3$ .

Checking the remaining cases by hand ( $n \le 6$ ), we find the couples $(2,1), (3,1), (5,2)$ .

The answer is $\boxed{15}$