Proving $n^{n+r}>(n+r)^n$ 
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 ?
 A: I would do something similar.
We want to show $n^r< (1+\frac{r}{n})^n$.
Now notice that $e^r=1+r+\frac{r^2}{2!}+\dots $
Using binomial expansion one can easily conclude that $ (1+\frac{r}{n})^n< e^r< n^r$ (because $\dfrac{\binom{n}{j}}{n^j}< \frac{1}{j!})$
A: Taking logarithms on both sides, the inequality is equivalent to
$$\frac{\ln(n + r)}{n + r} < \frac{\ln(n)}{n}.$$
Now consider the function $x \mapsto \frac{\ln(x)}{x}$. The derivative is $\frac{1 - \ln(x)}{x^2}$, which is negative for $x > e$. But this means that the function is decreasing in this range, which proves the original inequality.
A: $n^{n+r}\gt(n+r)^n$
$n^{n+r\over n }\gt(n+r)$
$n^{1\over n}\gt (n+r)^{1\over n+r}$
$^{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

A: Put $n+r=m$ and prove that
$n^m>m^n$ if $m\geq n+3$.
or $\frac{\ln(n)}{n}-\frac{\ln(m)}{m}>0$.
the derivative of the function $f: x\mapsto \frac{\ln(x)}{x}$ has the sign of the numerator $(1-\ln(x))$ so,
$f$ is decreasing at $[3,+\infty)$, and
$$\forall m>n>2 \;\;\frac{\ln(m)}{m}<\frac{\ln(n)}{n}$$ qed.
