New Year Enigma

This solution is a little on the long side but because this kind of problem appeals to a large grouping of people with varying abilities I felt it best to leave nothing unexplained. $A=\sqrt [ 2015 ]{ 2015! } \\ A^{ 2015 }=2015!\\ A^{ 2015 }\cdot A=2015!\cdot A\\ A^{ 2016 }=2015!\cdot \sqrt [ 2015 ]{ 2015! } \\ \\ B=\sqrt [ 2016 ]{ 2016! } \\ B^{ 2016 }=2016!\\ B^{ 2016 }=2015!\cdot 2016\\ \\ Because\quad we\quad know\quad A\quad and\quad B\quad are\quad positive,\quad we\quad can\quad say\\ 1)\quad Iff\quad B^{ 2016 }>A^{ 2016 }\quad then\quad B>A\\ 2)\quad Iff\quad A^{ 2016 }>B^{ 2016 }\quad then\quad A>B\\ 3)\quad Iff\quad A^{ 2016 }=B^{ 2016 }\quad then\quad A=B\\ \\ Because\quad A^{ 2016 }\quad and\quad B^{ 2016 }\quad can\quad both\quad be\quad written\quad as\quad 2015!\quad multiplied\\ by\quad some\quad constant,\quad whoever's\quad constant\quad is\quad greater\quad will\quad be\quad the\quad greater\quad number\\ \\ The\quad constant\quad which\quad when\quad multiplied\quad with\quad 2015!\quad gives\quad A^{ 2016 }\quad is\quad \sqrt [ 2015 ]{ 2015! } \\ \\ The\quad constant\quad which\quad when\quad multiplied\quad with\quad 2015!\quad gives\quad { B }^{ 2016 }\quad is\quad 2016\quad \\ \\ \sqrt [ 2015 ]{ 2015! } =\sqrt [ 2015 ]{ 2015\cdot 2014...\cdot 2\cdot 1 } \quad Note\quad that\quad within\quad the\quad 2015^{ th\quad }root\quad are\\ 2015\quad numbers,\quad all\quad of\quad which\quad are\quad less\quad than\quad 2016,\quad multiplied\quad together\\ \\ 2016=\sqrt [ 2015 ]{ 2016^{ 2015 } } \quad Note\quad that\quad within\quad the\quad 2015^{ th }\quad root\quad are\quad 2015\quad numbers,\\ all\quad of\quad which\quad are\quad equal\quad to\quad 2016,\quad multiplied\quad together.\\ \\ Obviously\quad 2015\quad numbers\quad all\quad equal\quad to\quad 2016\quad multiplied\quad together\quad is\quad greater\\ than\quad 2015\quad numbers\quad all\quad less\quad than\quad 2016\quad multiplied\quad together\\ \\ Therefore:\\ 2016>\quad \sqrt [ 2015 ]{ 2015! } \\ B^{ 2016 }>A^{ 2016 }\\ B>A\\ \boxed { \sqrt [ 2016 ]{ 2016! } >\sqrt [ 2015 ]{ 2015! } }$

Each of the two numbers is the geometrical mean of the first n natural numbers:

$\sqrt [ 2015 ]{ 2015! } =\sqrt [ 2015 ]{ 2015\cdot 2014\cdot 2013\cdot ...\cdot 1 }$

Note that there are 2015 numbers, so the index is 2015.

$\sqrt [ 2016 ]{ 2016! } =\sqrt [ 2016 ]{ 2016\cdot 2015\cdot 2014\cdot ...\cdot 1 }$

Note that there are 2016 numbers, so the index is 2016.

Analysing a few examples, we can infer that the geometrical mean g(n) of the first n natural numbers increases as n does:

$g(2)=\sqrt{ 2! } =\sqrt{ 2 }\cong 1.4$

$g(3)=\sqrt [ 3 ]{ 3! } =\sqrt [ 3 ]{ 6 }\cong 1.8$

$g(4)=\sqrt [ 4 ]{ 4! } =\sqrt [ 4 ]{ 24 }>\sqrt [ 4 ]{ 16 }\therefore g(4)>2$

Hence, it follows that $g(2016)>g(2015)\therefore \sqrt [ 2016 ]{ 2016! }>\sqrt [ 2015 ]{ 2015! }$ .

$n!=n(n-1)...1$

$(n+1)^n = (n+1)(n+1)..(n+1)$ ( $n$ times)

Since each term of the product $(n+1)^n$ is greater than each term in $n!$

We have $n!<(n+1)^n$ for all natural numbers $n$

Write $(n+1)=\frac{(n+1)!}{n!}$ and multiply $(n!)^n$ to both the sides

So $(n!)^{n+1}<((n+1)!)^{n}$

Raising both sides by power $\frac{1}{n(n+1)}$

$(n!)^{\frac{1}{n}} < ((n+1)!)^{\frac{1}{n+1}}$

Put n=2015 and we get $(2015!)^{\frac{1}{2015}} < (2016!)^{\frac{1}{2016}}$ which is the desired result

NOTE: How I got this solution? I worked the same method backwards assuming this inequality. You may assume the inequality the other way to and get a contradiction.

I used small numbers to justify a trend. Rearranging the question and replacing with variables to look like

$(x!)^{\frac{1}{x}}$

It's easy to just plug in small values. When $x=2$ , the answer can be estimated around 1.5 when simplified to $(2)^{\frac{1}{2}}$ .

Use the same method for 3 to compare a higher number that can still be estimated in your head. In this case it's simplified to $(6)^{\frac{1}{3}}$ which can be estimated to 1.8. It can now be said that $f(x) > f(x-1)$ . Not the best way to do it but it's an easy way if you have just a little experience with roots.

By Stirling's approximation ln (n!)= n ln n-n, i.e., n!= (n/e)^n. Therefore n^th root of n! is n/e. So the result now follows.

Only catch is Stirling's approximation is valid only for large n.