<s>[INST] Prove that $(n+1)^n < n^{n+1}$ for all $n>3$ 
Prove that $(n+1)^n < n^{n+1}$ for all $n>3$

At $n=4$, $$5^4<4^5$$  which is indeed true.
By mathematical induction, we need to prove that $$(n+2)^{n+1} < (n+1)^{n+2}$$
$$\implies (n+2)^n\times (n+2) < (n+1)^n\times(n+1)^2 $$
I am not getting how to proceed further than this. Any hints or help would be highly appreciated! Thanks.
 [/INST] Here is a non-inductive proof that involves a nice application of the AM-GM inequality.
It suffices for us to prove the following inequality:
\begin{align}
\forall \ & n>3,  \left(1+\dfrac{1}{n}\right)^n <4 \\
& \iff \left(\dfrac{n+1}{n}\right)^n <4 \\
& \iff \left(\dfrac{n}{n+1}\right)^n >\dfrac{1}{4} \\
& \iff \dfrac{n}{n+1} > \sqrt[^n]{\dfrac{1}{4}}
\end{align}
But
\begin{align}
\sqrt[^n]{\dfrac{1}{4}} &= \sqrt[^n]{\dfrac{1}{2} \cdot \dfrac{1}{2} \cdot 1^{n-2}} \\
& \leq \dfrac{\dfrac{1}{2}+ \dfrac{1}{2}+n-2}{n} (\text{By the AM-GM Inequality}) \\
& = \dfrac{n-1}{n}.
\end{align}
It remains to show that $\dfrac{n-1}{n} < \dfrac{n}{n+1}$, which is obvious since $n^2-1 < n^2$.
</s>