<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] Using binomial expansion, we can write
\begin{align*}
\text{LHS} &= (n+1)^n\\
&= (n+1)^{n-1}\times (n+1)\\
&= \left[n^{n-1} + {n-1 \choose 1}n^{n-2} + {n-1 \choose 2}n^{n-3} + \cdots + {n-1 \choose n-2}n + {n-1 \choose n-1}\right] \times (n+1)\\
&= \left[n^{n-1} + {n-1 \choose 1}n^{n-2} + {n-1 \choose 2}n^{n-3} + \cdots + {n-1 \choose n-2}n\right] \times (n+1) + (n+1)\\
\end{align*}
For $n > 3$, it is straightforward to show that
$${n-1 \choose k} < n^k$$
for all $k = 1,2,\dots, n-2$. Thus, we have
\begin{align*}
\text{LHS} &< \left[n^{n-1} + n^1n^{n-2} + n^2n^{n-3} + \cdots + n^{n-2}n\right] \times (n+1) + (n+1)\\
&= \left[(n-1)n^{n-1}\right] \times (n+1) + (n+1)\\
&= (n^2 - 1)n^{n-1} + (n+1)\\
&= n^{n+1} - n^{n-1} + (n+1)\\
&< n^{n+1} = \text{RHS},
\end{align*}
where the last step follows from the fact that
$$n^{n-1} > n+1$$
for all $n > 3$.
Remark: In contrast to other answers, this gives a proof without using induction.
</s>