Prove $n^{n+1}$ is greater than $(n+1)^n$ for all $n > 2$ The question of whether $2014^{2015}$ or $2015^{2014}$ is greater came up in my Calculus class and it seems clear that $2014^{2015} > 2015^{2014}$ based on a table comparing the sequences $s_n = n^{n+1}$ and $t_n = (n+1)^n$ for $n > 2$. However, I am unable to prove this result.
Thus far, I attempted an inductive proof without success as well as recasting the question as an analysis question where $f(x) = x^{x+1}$ and $g(x) = (x+1)^x$. It is easy to show that $f(3) > g(3)$, thus, if it could be shown that $f'(x) > g'(x)$ for all $x \geq 3$ the proof would follow easily. However, the derivatives are not `nice.' 
Any help would be appreciated! 

$n^{n+1}/(n+1)^n = n ( n^n / (n+1)^n) = n ( 1/(1+1/n)^n) \geq n * 1/e >1 $ for $n>2$ since $(1+1/n)^n \to e$ monotonically increasing. Thus, $n^{n+1} > (n+1)^n$. 
