find the least value of $n$ such $2017^{{2017}^{2017}}~~|~n!$ 
Let $n$ be positive integer. Find the minimum of the $n$ such
  $$2017^{{2017}^{2017}}~~|~n!$$

[Note that 2017 is a prime]
use this formula:
$$v_{2017}(n!)=\dfrac{n-S_{2017}(n)}{2016}$$so find least $n$ such $$ n-S_{2017}(n)\ge 2017^{2017}$$
$S_{p}(n)$ denotes the sum of the standard base-p digits of n, then How find this  least $n?$

idea
If $n=2017^{k}$, then
$$\nu_{2017}(n!)=\sum_{j=1}^{k}\left\lfloor\frac{n}{2017^j}\right\rfloor=1+\dotsb +2017^{k-1}=\frac{2017^k-1}{2016}$$
Now perhaps we want $k$ such that
$$\frac{2017^k-1}{2016} \geq 2017^{2017}.$$
Note: this is just an idea which may require a bit more massaging to get the least value of $n$.
