consider when $n<47$ $\underline{case1}:~n$ is a prime number thus $n\leq43<\sum_{i=1}^\infty \lfloor \frac{2016}{43^i}\rfloor\Rightarrow n<v_n(2016!)$ which means $n^n|2016!$, a contradiction $\underline{case2}:~n$ is a composite number therefore each prime factor of $n$ is less than or equal to $\sqrt{n}$ let $k$ be a prime factor of $n$ we'll get that $k\leq\sqrt{47}<7$ thus $v_k(2016!)=\sum_{i=1}^\infty \lfloor \frac{2016}{k^i}\rfloor\geq\sum_{i=1}^\infty \lfloor \frac{2016}{6^i}\rfloor>336$ since $k<43,2^6>43\Rightarrow v_k(n)\leq 5\Rightarrow v_k(n^n)\leq 5n<5(43)=215$ therefore $v_k(n^n)<v_k(2016!)$ for every prime factor of $n$ which means $n^n|2016!$, a contradiction $\Rightarrow n\geq47$ since $v_{47}(2016!)=\sum_{i=1}^\infty \lfloor \frac{2016}{47^i}\rfloor=42<47$ we'll get that $47^{47}\nmid 2016!$ $\therefore~min(n)=47~~~\square$