$n^n<(n-1)^{n+1}$ 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.

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 $n>4$.
