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.
 A: it is easier if you rearrange the statement a little. dividing throughout by the +ve quantity $n^n$ it becomes the claim that 
$$
\left(1+\frac1n\right)^n \lt n
$$
which is true if $n=3$.
now multiply both sides by the +ve quantity  $\left(1+\frac1n\right)$.
A: HINT for the induction step: If your induction hypothesis is that $n^{n+1}>(n+1)^n$, then
$$(n+1)^{n+2}=(n+1)n^{n+1}\left(\frac{n+1}n\right)^{n+1}>(n+1)^{n+1}\left(\frac{n+1}n\right)^{n+1}\;.\tag{1}$$
Now combine everything on the righthand end of $(1)$ into a single $(n+1)$-st power and do a little algebra.
A: Your inequality is the same as 
$$
n\left(\frac{n}{n+1}\right)^n\gt1
$$
Notice that
$$
\begin{align}
(n+1)\left(\frac{n+1}{n+2}\right)^{n+1}
&=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\
&=\left(1+\frac1{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\[9pt]
&\gt1\cdot1
\end{align}
$$
A: 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 $(n+1)^{n+2}>(n+2)^{n+1}$.
A: Assume the statement holds true for $n-1$, i.e. we have $$(n-1)^n>n^{n-1}.$$
Then 
\begin{align*}
(n+1)^n &= n^n+\binom{n}{1}n^{n-1}+\dots+\binom{n}{n-1}n+1\\
&<n^n+n^n+\dots+n^n=n(n^n)=n^{n+1}.
\end{align*}
