<s>[INST] 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$ 
 [/INST] 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.
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