Power of 2016 dividing 2016!

The index of a prime $p$ in $N!$ is $I_p(N) \; = \; \sum_{n=1}^\infty \left\lfloor \frac{N}{p^n}\right\rfloor$ Since $I_2(2016) = 2010$ , $I_3(2016) = 1004$ , $I_7(2016) = 334$ we deduce that $2016! \; = \; 2^{2010} \times 3^{1004} \times 7^{334} \times q$ where $q$ and $42$ are coprime. Now $2016 = 2^5 \times 3^2 \times 7$ , and hence $2016^n$ divides $2016!$ precisely when $5n \le 2010$ , $2n \le 1004$ and $n \le 334$ . Thus $2016^n$ divides $2016!$ when $n \le \boxed{334}$ .

Notice that from $1$ to $2016$ there are:

$\left\lfloor \frac { 2016 }{ 7 } \right\rfloor =288$ numbers that have at least one factor of $7$

$\left\lfloor \frac { 2016 }{ 49 } \right\rfloor =41$ more numbers that have an additional factor of $7$

$\left\lfloor \frac { 2016 }{ 343 } \right\rfloor =5$ more number that adds an additional factor of $7$

So, $n\le 288+41+5=334$ or ${ 2016 }^{ n }$ will have more factors of $7$ than ${ 2016 }!$ , which means that ${ 2016 }^{ n }$ cannot divides ${ 2016 }!$ .

$\therefore \quad { n }_{ max }=334$

$~~~~~\\ ~~~\\ \Large \text{These extra lines on the top, in black, are not mine!! it is not what I have written.}\\ ~~~\\ \color{#E81990} {2016! =32*9*7=2^5*3^2*7.}\\ ~~\\ \color{#20A900} {Number~ of~2^5~in~2016!\\ \left\lfloor \frac { 2016 }{ 2 } \right\rfloor =1008,~at~least~1008~ numbers~ that~ has~ 2 ~as~ factor.......\\ \left\lfloor \frac { 2016 }{ 2^2 } \right\rfloor =~~ 504~ numbers~ that~ has~ 2 ~as~ factor,~additional~number~since~4~is~a~factor.......\\ \left\lfloor \frac { 2016 }{ 2^3 } \right\rfloor =~~ 252~ numbers~ that~ has~ 2 ~as~ factor,~additional~number~since~8~is~a~factor.......\\ \left\lfloor \frac { 2016 }{ 2^4} \right\rfloor =~~ 101~ numbers~ that~ has~ 2 ~as~ factor,~additional~number~since~16~is~a~factor.......\\ 1008+504+252=1764,~that~is~~2^{1764}>=(2^5)^{352}. ~In ~fact~there~are~more ~2s~due~to~higher~powers~of~2.\\ So~2016!~has~more~than~352~~~"~2^5=32~".}\\ ~~~~\\ \color{#EC7300} {Number~ of~3^2~in~2016!\\ \left\lfloor \frac { 2016 }{ 3 } \right\rfloor =672,~~at~least~672~ numbers~ that~ has~ 3 ~as~ factor.......\\ 3^{672}=(3^2)^{336}.~~~In ~fact~there~are~more ~3s~due~to~higher~powers~of~3.\\ \\ So~2016!~has~at~least~~352~~"~3^2=9~".}\\ ~~~~\\ \color{#3D99F6} {Number~ of~7~in~2016!\\ \left\lfloor \frac { 2016 }{ 7 } \right\rfloor =288.~ ~at~least~288~ numbers~ numbers~ that~ has~ 7 ~as~ factor........\\ \left\lfloor \frac { 2016 }{ 7^2 } \right\rfloor =~41~ numbers~ that~ has~ 7 ~as~ factor,~additional~number~since~7^2~is~a~factor.......\\ \left\lfloor \frac { 2016 }{ 7^3} \right\rfloor =~5~ numbers~ that~ has~ 7 ~as~ factor,~additional~ number~since~ 7^3~is~a~ factor....\\ Since~7^4>2016, ~no~more~of~7~is~available.\\ 288+41+5=334.}\\ ~~~\\ Thus ~only~334~of~7~are~available.\\ So~n=\Large \color{#D61F06}{334}.$

Thanks to Mr. Linkin Duck . I adopted his style of presentation. As such I have only added more explanation to his solution.