if $n>3$ and $n+1$ was composite then we would have $n+1 \mid n!$ so we get that $n+1 \mid 1$... so we have $n\leq 3$ or $n+1$ is a prime number... so assume that $n+1$ is prime,then we have $n!=(n+1)^{k}-1=n[(n+1)^{k-1}+...+1]$ so we get that $(n-1)!=(n+1)^{k-1}+...+1$ now if $n>4$ then its a composite number,thus we would have: $(n+1)^{k-1}+...+1 \equiv k (\bmod n)$ and also $n \mid (n-1)!$ so we must have $n \mid k$ so $k\geq n$ but one can see that if $n\geq 2$: $(n+1)^{k}\geq (n+1)^{n}>(n-1)!+1$ so in this case we wouldnt have any solutions... so the remain case is $n\leq 4$ which would give us $n=4,k=2$ and $n=2,k=1$...

BaBaK Ghalebi wrote: if $n>3$ and $n+1$ was composite then we would have $n+1 \mid n!$ so we get that $n+1 \mid 1$. Can you explain this further?

maybe the folowing explanation would make it clear... well if $n+1$ was composite then we would have $n+1=pq$ s.t $\gcd (p,q)=1$ and $p,q>1$ so we get that $p,q<n+1$ so $p,q\leq n$ so we get that: $i)p \mid n!$ $ii)q \mid n!$ now from $i),ii)$ and $\gcd (p,q)=1$ we get that $pq \mid n!$ so $n+1 \mid n!$... now in the equation $(n+1)^{k}-1=n!$ we have $n+1 \mid RHS$ so we get that $n+1 \mid LHS$ so $n+1 \mid 1$...

BaBaK Ghalebi wrote: well if $n+1$ was composite then assume that $p^{\alpha}\mid n+1$ s.t $\alpha>1$ Alright; everything made sense except for why you can assume that $\alpha >1$. It doesn't seem like this treats the case of something like, $n+1=6=2 \cdot 3$. Then, you wouldn't have $2p \leq p^{\alpha-1}<n \iff 4 \leq 1 <5$. Unless I'm missing something, it doesn't look like you treat the cases where $n+1 = \prod p_{i}$; only cases where $n+1$ is prime or is composed of primes with higher powers than $1$.

cincodemayo5590 wrote: BaBaK Ghalebi wrote: well if $n+1$ was composite then assume that $p^{\alpha}\mid n+1$ s.t $\alpha>1$ Alright; everything made sense except for why you can assume that $\alpha >1$. It doesn't seem like this treats the case of something like, $n+1=6=2 \cdot 3$. Then, you wouldn't have $2p \leq p^{\alpha-1}<n \iff 4 \leq 1 <5$. Unless I'm missing something, it doesn't look like you treat the cases where $n+1 = \prod p_{i}$; only cases where $n+1$ is prime or is composed of primes with higher powers than $1$. sorry it was a typo... I edited my post,I think its correct now...

BaBaK Ghalebi wrote: cincodemayo5590 wrote: BaBaK Ghalebi wrote: well if $n+1$ was composite then assume that $p^{\alpha}\mid n+1$ s.t $\alpha>1$ Alright; everything made sense except for why you can assume that $\alpha >1$. It doesn't seem like this treats the case of something like, $n+1=6=2 \cdot 3$. Then, you wouldn't have $2p \leq p^{\alpha-1}<n \iff 4 \leq 1 <5$. Unless I'm missing something, it doesn't look like you treat the cases where $n+1 = \prod p_{i}$; only cases where $n+1$ is prime or is composed of primes with higher powers than $1$. sorry it was a typo... I edited my post,I think its correct now... OK, that makes more sense now. Thanks.

Everything is fine until here BaBaK Ghalebi wrote: now if $n>4$ then its a composite number,thus we would have: $(n+1)^{k-1}+...+1 \equiv k (\bmod n)$ and also $n \mid (n-1)!$ so we must have $n \mid k$ so $k\geq n$ but one can see that if $n\geq 2$: $(n+1)^{k}\geq (n+1)^{n}>(n-1)!+1$ so in this case we wouldnt have any solutions... so the remain case is $n\leq 4$ which would give us $n=4,k=2$ and $n=2,k=1$... Then I think your english gets in the way of the rest of the proof, at least for me...

Well yeah, my proof is pretty much the same. For $n>4$ it is clear that $n$ muc be even so let $n=2m$ and so $n!= (2m)! = (2m).(2m-1)!$ so it is clear that $n^{2}| n!$. So $n^{2}| (n+1)^{k}-1$. Using binomial expansion we see that if $n^{2}| (n+1)^{k}-1$ then $n|k$ and so $k \geq n$. Therefore $(n+1)^{k}\geq (n+1)^{n}> (n-1)!+1$. Now all we have to do is check for $n=1,2,3,4$.

me@home wrote: Everything is fine until here BaBaK Ghalebi wrote: now if $n>4$ then its a composite number,thus we would have: $(n+1)^{k-1}+...+1 \equiv k (\bmod n)$ and also $n \mid (n-1)!$ so we must have $n \mid k$ so $k\geq n$ but one can see that if $n\geq 2$: $(n+1)^{k}\geq (n+1)^{n}>(n-1)!+1$ so in this case we wouldnt have any solutions... so the remain case is $n\leq 4$ which would give us $n=4,k=2$ and $n=2,k=1$... Then I think your english gets in the way of the rest of the proof, at least for me... well we assumed that $n+1$ is prime and also if $n>4$ then $n$ must be composite (because there doesnt exist two consecutive prime numbers greater than $4$) so $n$ is a composite number so $n \mid (n-1)!$ now from the equation $(n-1)!=(n+1)^{k-1}+...+1$ we get that $n \mid LHS$ so $n \mid RHS$ but $RHS \equiv 1+1+...+1 \equiv k (\bmod n)$ so we get that $n \mid k$ so $n\leq k$,but we know that the folowing inequality is true for $n\geq 2$: $(n+1)^{k}\geq (n+1)^{n}>(n-1)!+1$ contradicting the equality of the problem... so we have $n\leq 4$ so we only need to check these cases... I hope its clear now...