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 $n \ge 4$)

Rearrange the inequality a little:
$$\left(\frac{n-1}{n}\right)^n > \frac1n.$$
Then you get
$$\begin{align}
\left(\frac{n}{n+1}\right)^{n+1} &= \left(\frac{n}{n+1}\right)^{n+1} \left(\frac{n}{n-1}\right)^n \left(\frac{n-1}{n}\right)^n\\
&= \frac{n}{n+1} \left(\frac{n^2}{n^2-1}\right)^n\left(\frac{n-1}{n}\right)^n\\
&> \frac{n}{n+1} \left(\frac{n^2}{n^2-1}\right)^n \frac1n\\
&= \frac{1}{n+1} \left(\frac{n^2}{n^2-1}\right)^n\\
&> \frac{1}{n+1}.
\end{align}$$
