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

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 $n \ge 5$.
