Prove inequality by induction $ n^n >(n+1)^{n-1}$

Question

How to prove for $n\ge 2$ the following inequality by induction? I have no idea how to do it. $$ n^n >(n+1)^{n-1}$$ I know that inductive step is $ (n+1)^{n+1} >(n+2)^{n}$, but what next?

Answer 1

$$(n+1)^{n+1}=n^n\cdot\left(1+{1\over n}\right)^n(n+1)>(n+1)^{n-1}(n+1)\left(1+{1\over n}\right)^n$$

$$=(n+1)^n\left(1+{1\over n}\right)^n=\left(n+{1\over n}+2\right)^n>(n+2)^n$$

Answer 2

Why use induction? Here's an easier approach: the RHS can be rewritten as $n \log n +n \log (1+\frac{1}{n}) - \log n - \log (1+\frac{1}{n})$. After cancellation and putting $\log n$ on LHS you can expand the remaining terms in Maclaurin series and it's easy to see that LHS grows as $\log n$ and RHS converges to 1.

Answer 3

With inductions like these where it's not obvious, we can try 'forcing in' the previous step and see what happens:

$$(n+1)^{(n+1)} = {(n+1)^{(n+1)}n^n\over n^n}$$

Now we can use the inductive step.

$${(n+1)^{(n+1)}n^n\over n^n}>{(n+1)^{(n+1)}(n+1)^{(n-1)}\over n^n}$$ $$ = {(n+1)^{2n}\over n^n}$$

Now let's check this is greater than the RHS we want:

$${(n+1)^{2n}\over n^n} > (n+2)^n$$ $$\Leftarrow {(n+1)^2 \over n} > n+2$$ $$\Leftarrow (n+1)^2 > n(n+1)$$ Which is true for all n.

Answer 4

It holds for $n=1$ as $2^2>3^1$.

Assume it holds for $n=n$ then, \begin{align} &n^n>(n+1)^{n-1}\\ &n>\left(\frac{n+1}{n}\right)^{n-1}\\ &\frac{n+1}{n}\cdot n>\frac{n+1}{n}\cdot \left(\frac{n+1}{n}\right)^{n-1}\\ &n+1>\left(\frac{n+1}{n}\right)^{n}\\ \end{align} As $\frac{n+1}{n}>\frac{n+2}{n+1}$, \begin{align} &n+1>\left(\frac{n+1}{n}\right)^{n}>\left(\frac{n+2}{n+1}\right)^{n}\\ &\therefore (n+1)^{n+1}>(n+2)^n \end{align}