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 $n?$

Let $n=2016 \times 2017^{2017}+2017 $.
$S_{2017}(n)=2016+1=2017$,
$n-S_{2017}(n)=2016 \times 2017^{2017}+2017-2017=2016 \times 2017^{2017}$,
$$v_{2017}(n!)=\dfrac{n-S_{2017}(n)}{2016}=\dfrac{2016 \times 2017^{2017}}{2016}=2017^{2017}$$
It won't be hard to prove this $n$ is optimal.
