Solve $1^n+2^n+ \ldots +n^n=k!$ over positive integers If $k,n \in \mathbb{N}^*$, solve the following equation
$1^n+2^n+ \ldots +n^n=k!$, where $k! $ denotes $1 \cdot 2 \cdot 3 \cdots k$.

Partial answer, complete for $n$ even.
Suppose $n\geq2$. Clearly, $k>n$.
Then $k!$ is even, so there is an even number of odd numbers on the left. So $n\equiv0,3\pmod4$.
Suppose $n\geq4$ is even. Odd squares are $1$ mod $8$, and $8\mid k!$, so $n/2$ is divisible by $8$. Odd $16$th powers are $1$ mod $32$ and $32\mid k!$, so $n/2$ is divisible by $32$. [...] Odd $4^m$th powers are $1$ mod $2^{2m+1}$ (Euler) so $2^{2m+1}\mid n/2$... etc, contradiction. (Boring details left out.)
Suppose $n\geq3$ is odd. Odd numbers have odd powers that are congruent to themselves mod $8$ and $8\mid k!$, so $8$ divides $1+3+\cdots+n=(n+1)^2/4$, so $n\equiv-1\pmod{8}$. Then odd numbers have $n$th powers that are congruent to their inverse modulo $32$ and $32\mid k!$, so $32$ divides $(n+1)^2/4$, so $n\equiv-1\pmod{16}$. From now this argument does not give anything new: $64\mid(n+1)^2/4$ only implies $n\equiv-1\pmod{16}$.
