Prove that $(n+2)^n < (n+1)^{n+1}$ for all $n \in \Bbb N$

Question

Is there a simple(ish) way of proving this? I got to this step when I was trying to show that the sequence $a_n = (1 + \frac{1}{n})^n$ is increasing for all $n \in \Bbb N$. It came up after I expanded out each term using the binomial theorem, and then tried to find the $n$ that makes $a_{n+1} > a_n$ for each term.

Prove that $(n+2)^n < (n+1)^{n+1}$ for all $n \in \Bbb N$

Answer 1

Have you tried induction?

For $n=0$ this is wrong, so you dont consider $0\in\mathbb{N}$?

For $n=1$ we have $3<4\checkmark$

The Inductive claim:

(I.C.)For arbitrary $n\in\mathbb{N}$ holds $(n+2)^n<(n+1)^{n+1}$.

For the inductive step:

$(n+2)^{n+2}=(n+2)(n+2)^{n+1}>(n+2)(n+1)^{n+1}\stackrel{I.C.}{>}(n+2)(n+2)^n=(n+2)^{n+1}$

Answer 2

By AM-GM we have:$$(n+2)^{n}\cdot 1 < \left(\dfrac{(n+2)n+1}{n+1}\right)^{n+1}$$