How to prove this $(n+1)^n < n^{n+1}$ for $\space n \ge 3$ I'm having some more trouble with induction
I know how to prove this using  $\ln$, but I need to use induction only.
prove that:
$(n+1)^n < n^{n+1}$  
for any $ n\ge  3$ 
 A: You need only to show:
$$\frac{(n+1)^n}{n^{n+1}}>\frac{(n+2)^{n+1}}{(n+1)^{n+2}}.$$
But this follows from
$$
(n+1)^{2n+2}=(n^2+2n+1)^{n+1}>(n(n+2))^{n+1}.
$$
A: Hypothesis
\[ n^{n+1}>(n+1)^{n},\ \mbox{for}\ n\ge 3\]
Basis
\[ 3^{3+1} > (3+1)^{3} \]
\[ 3^{4} > 4^{3} \]
\[ 81 > 64\]
Induction
\[
(n+1)^{n+2}=n^{n+1}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right]
\]
Now via the induction hypothesis, we have
\[
n^{n+1}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right] > (n+1)^{n}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right]
\]
$$
\begin{align}
(n+1)^{n}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right] &= (n+2)^{n+1}\left[\frac{(n+1)^{2n+2}}{n^{n+1}(n+2)^{n+1}}\right]\\&= (n+2)^{n+1}\left[\frac{(n^{2}+2n+1)^{n+1}}{(n^{2}+2n)^{n+1}}\right]
\end{align}
$$
Since $n$ is positive and the numerator is larger than the denominator, then the fraction is greater than $1$. Hence,
\[
(n+2)^{n+1}\left[\frac{(n^{2}+2n+1)^{n+1}}{(n^{2}+2n)^{n+1}}\right] > (n+2)^{n+1}
\]
Thus,
\[ n^{n+1}>(n+1)^{n},\ \mbox{for}\ n\ge 3\]
I hope this helps you understand.
A: You can also use the Bernoulli inequality $1+na\leq (1+a)^n$ which is valid for $-1\leq a$
write your inequality as
$$\frac{n+1}{n^2} < \frac{n^{n-1}}{(n+1)^{n-1}}=\left(1-\frac{1}{n+1}\right)^{n-1}$$
now by Bernoulli
$$\frac{2}{n+1} =1-\frac{n-1}{n+1}\leq \left(1-\frac{1}{n+1}\right)^{n-1}$$
so it remains to show 
$$\frac{n+1}{n^2} <\frac{2}{n+1}, $$
which is equivalent with $2< (n-1)^2$ and this is true for $n\geq 3$.
A: Taking natural logarithm, you will get
$$ \frac{\ln(n+1)}{n+1}<\frac{\ln n}{n}. \tag{1}$$
Let $f(x)=\frac{\ln x}{x}$ for $x\ge 1$. Then $f'(x)=\frac{1-\ln x}{x^2}<0$ for $x>1$, namely $f(n)$ is decreasing for $n>1$. Thus $f(n+1)<f(n)$ which is (1). 
