Find all integer $(n,k)$ such that $n!=(n+1)^k-1$
My Attempt . Let $(n+1)=p$. Therefore,$ L.H.S. \equiv {-1}\pmod p$ and $R.H.S. \equiv {-1}\pmod p$. So,$k=log_{p} ^{(p-1)!+1}$. Now we have to set the values of $p$ in order to find integer $k$. I am not satisfied of my answer.I believe that better answer is there.I didn't complete my answer because I considered a certain case(i.e. $n+1$ is a prime).I must consider more cases but i failed to find them! Please help me!
Suppose that $n>2$ and $k>0$.
$n^k \leq (n+1)^k-1=n!<n^n$ so that $k<n$.
If $n=ab$ for some $a \neq b$, then $n!$ contains the terms $a,b,ab$ in the product and hence is divisible by $n^2$. OTOH the R.H.S. is equal to $kn$ modulo $n^2$, that is: $n$ divides $k$, contradicting 1.
Unless $n$ is of the form $p^2$, $n$ has a representation as in 2. Since $n$ is even, for $n>4$, $n$ is not of the form $p^2$ and thus there is no solution for $n>4$.