<s>[INST] Induction: $n^{n+1} > (n+1)^n$ and $(n!)^2 \leq \left(\frac{(n + 1)(2n + 1)}{6}\right)^n$ How do I prove this by induction:
$$\displaystyle n^{n+1} > (n+1)^n,\; \mbox{ for } n\geq 3$$
Thanks.
What I'm doing is bunch of these induction problems for my first year math studies. 
I tried using Bernoulli's inequality at some point, but no success. Also, tried $(n+1)^{n+2}=(n+1)^{n+1}(n+1)$, then expanding $(n+1)^{n+1}$ by binomial formula to get the $n^{n+1}$ member to apply the induction hypothesis, still no success.
Here's another one I've been struggling with:
$$(n!)^2 \leq \left(\frac{(n + 1)(2n + 1)}{6}\right)^n$$
EDIT: Finally solved the second one!
What I needed was the AM-GM inequality.
Therefore,
$$\frac{(n + 1)(2n + 1)}{6} = \frac{1}{n} \sum_{i=1}^{n} i^2 \geq \sqrt[n]{1^2 \cdot 2^2 \cdots n^2}$$
Thus,
$$\left(\frac{(n + 1)(2n + 1)}{6}\right)^n = \left(\frac{1}{n} \sum_{i=1}^{n} i^2\right)^n \geq 1^2 \cdot 2^2 \cdots n^2 = (n!)^2$$
Done.
 [/INST] Your inequality is the same as 
$$
n\left(\frac{n}{n+1}\right)^n\gt1
$$
Notice that
$$
\begin{align}
(n+1)\left(\frac{n+1}{n+2}\right)^{n+1}
&=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\
&=\left(1+\frac1{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\[9pt]
&\gt1\cdot1
\end{align}
$$
</s>