# Find all $n,k \in \Bbb N^+$ such that $(n-1)!=n^k-1$ [duplicate]

Find all $$n,k \in \Bbb N^+$$ such that $$(n-1)!=n^k-1$$

Using $$n^k-1=(n-1)(1+n^2+...+n^{k-1})$$ and cancelling $$(n-1)$$ in each side, we get that $$(n-2)!|1+n^2+...+n^{k-1}$$ So, let's suppose $$n$$ is an even number, in $$(1+n^2+...+n^{k-1})$$ we have $$k$$ terms, where $$k-1$$ are even ($$n^i$$ for $$i>0$$) and one is odd ($$1$$), so we have that $$1+n^2+...+n^{k-1}$$ is odd. Also, if $$n$$ is even, then $$n-2$$ has the same parity so n-2 is even; therefore, if $$n-2>0$$ then $$(n-2)!$$ is even and that is a contradiction. So with this, when n is an even number, the only solution is $$n=2, k=1$$ Now, in every other solution, $$n$$ is odd.

Let's suppose $$k=1$$. we have $$(n-2)!=1$$ (for $$n, n-2$$ odd) and it follows $$n=3, k=1$$.

Using this, if $$n$$ is odd, then $$n^x$$ is odd too, so it follows that we have $$k$$ odd terms, so if $$k$$ is even, then $$(1+n^2+...+n^{k-1})$$ is even, since every pair of terms sums an even number. From $$n=4$$ onwards, $$(n-2)!$$ is even so k has to be even too.

Let's suppose $$k=2$$. Then $$(n-2)!|n^2+1$$ $$n=5$$ works and let's prove is the only solution. Supose there is a $$n=5+m$$ for $$m>0$$. So: $$(m+3)!|m+6$$ And it's not hard to prove that $$(m+3)!>m+6$$ if $$m$$ is not $$0$$

So I got stuck here, I could only find these 3 solutions, and I don't know how to prove that there aren't more, since, for more bigger values it seems that $$(n-2)!$$ is greater if $$k. Would appreciate any help on the proof, or hints. thanks.

• Let $p$ be a prime factor of $n$. Since $p$ does not divide $n^k-1$, $p$ does not divide $(n-1)!$, so $p \geq n$. So $n$ is prime. If $n$ is prime, by Euler theorem there is $\omega_n > 0$ such that $(n-1)!|n^k-1$ iff $\omega_n|k$. Sep 11, 2019 at 21:56
• SonodaUmi, I found (n=3, ∀k ∈N),(n=4, k=5 ), (k=6, has no solution for k), (n=7, k must be found by brute force). I concluded that there are always solutions for any n except those their last digit is 6 or 8. Sep 12, 2019 at 19:24
• I'm sorry, i had an error on the title. I'll correct it now Sep 12, 2019 at 22:28
• It was suposed to be equal, not only divisible, sorry Sep 12, 2019 at 22:30
• Where does this problem come from? Sep 13, 2019 at 2:51