proof that the inequality $n (n-1)^n > n^n$ holds for all $n \ge 4$ I am trying to prove that the inequality $n(n-1)^n > n^n$ holds for all $n\ge4$.
I tried using mathematical induction, but I really couldn´t find a way how to get past the $P(n) \implies P(n+1)$ step. I get $n(n-1)^n > n^n$ implies $n^{n+1} > (n+1)^n$   I tried expanding with binomial theorem but to no avail. I also know that in the limit the equality holds (but I want to prove it for all $n \ge 4$)
 A: Equivalently, we want to prove that $(n-1)^n \gt n^{n-1}$. Taking logarithms, we see that we want to prove that $n\log(n-1)\gt (n-1)\log n$, or equivalently that 
$$\frac{\log(n-1)}{n-1}\gt \frac{\log n}{n}.$$
Consider the function
$$f(x)=\frac{\log x}{x}.$$
We have 
$$f'(x)=\frac{1-\log x}{x^2}.$$
The derivative $f'(x)$ is negative for $x\gt e$. So $f(x)$ is decreasing in the interval $[e,\infty)$.
In particular, if $n\ge 4$, then $f(n-1)\gt f(n)$.
A: Your inequality is equivalent$^{1}$ to $$n+1 > {\left( {1 + \frac{1}{n}} \right)^{n + 1}}$$ for $n\geqslant 3$. What do you know about the sequence ${\left( {1 + \dfrac{1}{n}} \right)^{n + 1}}$ that might help you see why this is true?
$1.$ $$\begin{align}
  n{\left( {n - 1} \right)^n} > {n^n} \cr 
  \left( {n + 1} \right){n^{n + 1}} > {\left( {n + 1} \right)^{n + 1}} \cr 
  n + 1 > {\left( {\frac{{n + 1}}{n}} \right)^{n + 1}} \cr 
  n + 1 > {\left( {1 + \frac{1}{n}} \right)^{n + 1}} \end{align} $$
A: Rearrange the inequality a little:
$$\left(\frac{n-1}{n}\right)^n > \frac1n.$$
Then you get
$$\begin{align}
\left(\frac{n}{n+1}\right)^{n+1} &= \left(\frac{n}{n+1}\right)^{n+1} \left(\frac{n}{n-1}\right)^n \left(\frac{n-1}{n}\right)^n\\
&= \frac{n}{n+1} \left(\frac{n^2}{n^2-1}\right)^n\left(\frac{n-1}{n}\right)^n\\
&> \frac{n}{n+1} \left(\frac{n^2}{n^2-1}\right)^n \frac1n\\
&= \frac{1}{n+1} \left(\frac{n^2}{n^2-1}\right)^n\\
&> \frac{1}{n+1}.
\end{align}$$
A: Suppose $n-1\geq e$. Then $$n \geq \frac{n}{n-1}e > \left(1+\frac{1}{n-1}\right)^n$$
which rearranges to your inequality. Since $e+1<4$, the result follows.
Note: We have used the $(n-1)$st convergent to $e$ to establish the second inequality.
