$n^n<(n-1)^{n+1}$ for integer $n\ge5$ 
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.
 A: I'd write $m=n-1$. Then the statement reduces to
$$\left(1+\frac1m\right)^m<\frac{m^2}{m+1}.$$
It's well-known that $(1+1/m)^m$ increases to $e$, but more naively,
$$\left(1+\frac1m\right)^m=1+1+\frac1{m^2}{m\choose 2}
+\frac1{m^3}{m\choose 3}+\cdots<1+1+\frac12+\frac16+\cdots<3.$$
But
$$\frac{m^2}{m+1}>\frac{m^2-1}{m+1}=m-1>3$$
if $n>4$.
A: Here is a proof without induction. 
$(1-1/n)^n$ is strictly increasing for $n > 1 $ and approaches $1/e$ as $n \to \infty$ whereas $1/(n-1)$ is strictly decreasing and approaches $0$ as $n \to \infty$. Hence there is an $N_0$ such that for all $n > N_0$, 
$$
\Big(1 - \frac{1}{n}\Big)^n > \frac{1}{n-1}
$$
By little computation, we find that smallest $n$ for which this inequality holds is $n = 5$. Since LHS is increasing and RHS is decreasing, it implies that the inequality will never be violated for $n \ge 5$.
A: 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 $x\ge 5$.
