<s>[INST] 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 $1 \cdot 2 \cdot 3 \cdots k$.
 [/INST] This is a supplement to the answer by knm to show that $k\ge n+2$ (for large enough $n$).
Let $S_n=1^n+\cdots+n^n$. First, we show that $S_{n+1}/S_n>2(n+1)$. To do that, it suffices to show that $(r+1)^{r+1}/r^r>2(n+1)$ for all $r=0,\ldots,n$. Since the LHS is an increasing function, it has it's maximum at $r=n$, so we actually get
$$
\frac{(r+1)^{r+1}}{r^r} \ge \frac{(n+1)^{n+1}}{n^n}
> (n+1)\cdot\left(1+\frac{1}{n}\right)^n \ge 2(n+1).
$$
From this follows that
$$
\frac{S_{n+1}}{S_n} = \frac{\sum_{r=0}^n (r+1)^{r+1}}{\sum_{r=0}^n r^{r}}
> (n+1)\cdot\left(1+\frac{1}{n}\right)^n \ge 2(n+1).
$$
If $S_n>(n+1)!$ for some $n$, then if $S_n=k!$ we must have $k\ge n+2$.
However, if $S_n>(n+1)!$, then $S_{n+1}>2(n+1)S_n>(n+2)!$, so once we find one $n$ with $S_n>(n+1)!$, this is also the case for all larger $n$.
In particular, $S_3=36>4!$, so for all $n\ge3$ we have $S_n>(n+1)!$, which implies that $k\ge n+2$ if $k!=S_n$.
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