Prove that $n^{(n+1)} > (n+1)^{n}$ when $n\geq 3$ Firstly, for $n=3$,
$$3^4 > 4^3$$.
Secondly, $(n+1)^{(n+2)} > (n+2)^{(n+1)}$
Now I'm stuck. 
 A: Let's generalize $1$ to $k$, and consider the expression $$\frac{(n+k)^n}{n^{n+1}}=\left(\frac{n+k}{n}\right)^n\frac{1}{n}=\left(1+\frac{k}{n}\right)^n\frac{1}{n}$$
The first term, i.e. $\left(1+\frac{k}{n}\right)^n$, is an expression that is strictly increasing with $n$, and the limit is $e^k$.  Hence the expression will be less than 1 whenever $n>e^k$.  In the OP, $k=1$, so the expression will be less than 1 whenever $n>e$, i.e. $n\ge 3$.
A: A proof without induction  : 
Consider the function $f(x)=x^{\frac{1}{x}}$ .
The problem is equivalent with $f(n)>f(n+1)$ .
It's easy to find the derivative of $f$ :
$$f'(x)=x^{\frac{1}{x}-2}(1-\ln x)$$ so $f$ is strictly decreasing for every $x>e$ and so the conclusion holds (because $e<3$) :
$$f(n)>f(n+1)$$ which means that :
$$n^{(n+1)}>(n+1)^n$$
A: The inequality at stake is equivalent to $n > \left( 1+\frac{1}{n}\right)^n$.
Taking log yields $\frac{\ln n}{n} >\ln \left(1+\frac{1}{n} \right)$.
This is true since the concavity of $\log$ implies the stronger $\frac{1}{n} >\ln \left(1+\frac{1}{n} \right)$.
Using this argument, your inequality can be refined to $(n+1)^n<e^1n^n$
A: That is equivalent to proving that $a_n = \left(1+\frac{1}{n}\right)^n$ is an increasing sequence.
That follows, in a very straightforward way, from the AM-GM inequality, since:
$$\left(1+\frac{1}{n}\right)^n\cdot 1\leq\left(\frac{1+n\cdot\left(1+\frac{1}{n}\right)}{n+1}\right)^{n+1}=\left(1+\frac{1}{n+1}\right)^{n+1}. $$
