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.

Supposed for $n\geq 3$ you have that $n^{n+1}>(n+1)^n$. WTS $(n+1)^{n+2}>(n+2)^{n+1}$. Since $n^{n+1}>(n+1)^n$ you get that $n^{n+1} \cdot \frac{(n+1)^{n+2}}{n^{n+1}}>(n+1)^n\cdot \frac{(n+1)^{n+2}}{n^{n+1}}=\frac{(n+1)^{2n+2}}{n^{n+1}}>(n+2)^{n+1}$, where the last inequality follows since $(n+1)^2>n(n+2)$. Hence $(n+1)^{n+2}>(n+2)^{n+1}$.
