TITLE: Proving $n^{n+r}>(n+r)^n$
QUESTION [8 upvotes]: Proving $n^{n+r}>(n+r)^n$ for $n,r>2$, both natural numbers

I know that $n^{n+1}>(n+1)^n$, so if I only increase $''n+1''$ part by $1$, the inequality is still true, i.e. $n^{n+2}>(n+2)^n$ because the ratio of the left side is $n$ and the right side is $(1+\frac1{n+1})^n<e$, so iterating in this way I can reach $n+r$, do you agree ?

REPLY [2 votes]: $n^{n+r}\gt(n+r)^n$
$n^{n+r\over n }\gt(n+r)$
$n^{1\over n}\gt (n+r)^{1\over n+r}$
$^{n}\sqrt{n} \gt $ $^{n+r}\sqrt{n+r}$
we know that $^{x}\sqrt{x}$ is Descending after $e$
so if we increase x decrease value of function 
and we have $f'(x)=x^{{1\over x}-2}(ln x -1)$ so for x>2 derivative is negative and function is Descending