TITLE: $n^n<(n-1)^{n+1}$ for integer $n\ge5$
QUESTION [5 upvotes]: For natural number $n\ge5$, by mathematical induction or otherwise,
  prove that $n^n<(n-1)^{n+1}$.

Actually I was trying to solve the problem that I posted.

My Attempt:
Step I: Verify that when $n=5$, then the given inequality holds true;
$5^5\overset{?}{<}4^6\Rightarrow 3125<4096$ which is true.
Step II: Assume that the inequality is true for $n=k\ge 5$;
that is $k^k<(k-1)^{k+1}$ is true for $k\ge5$.
Step III: We need to prove that if the inequality is true for $n=k$, then it is true for $n=k+1$ as well;
that is to show that if $k^k<(k-1)^{k+1}$ is true, then $(k+1)^{k+1}<k^{k+2}$ is also true.
I have a difficulty to complete this part of the proof. Any help would be appreciated.

REPLY [3 votes]: The claim $x^x<(x-1)^{x+1}$ is equivalent to 
$$
x\ln x<(x+1)\ln(x-1).
$$
Consider $f(x)=(x+1)\ln(x-1)-x\ln x$ and prove that

$f'(x)=\ln\left(1-\frac1x\right)+\frac2{x-1}\to 0$ as $x\to +\infty$,
$f''(x)=-\frac{x+1}{x(x-1)^2}<0$ for $x>1$.

Then $f'(x)>0$ and $f$ increases. From $f(5)>0$ it follows that $f(x)>0$ for all $x\ge 5$.