Seeking a combinatorial proof ${(n+1)}^{n+1}>{(n+2)}^n$ I would appreciate if somebody could help me with the following problem
Q: Seeking a combinatorial proof that for all $n\in \mathbb{N}$, following holds
$${(n+1)}^{n+1}>{(n+2)}^n$$
 A: $(n+2)^n$ is the number of $n$-tuples of elements $\in\{0,1,\ldots,n+1\}$. To any such $n$-tuple we can assign an $(n+1)$-tuple of elements $\in\{0,1,\ldots,n\}$ in a way that allows us to reconstruct the input.
Let $(a_1,\ldots,a_n)$ be the given $n$-tuple. At least one of the $n+1$ values $0,\ldots,n$  does not occur among the $a_i$. Pick such a value $k$. For $1\le i\le n$ let $b_i=\begin{cases}a_i&\text{if $a_i<k$}\\a_i-1&\text{if $a_i>k$}\end{cases}$ and finally let $b_{n+1}=k$. This gives us an $(n+1)$-tuple with $b_i\in\{0,\ldots,n\}$ as desired.
To reconstruct the $a_i$ note that we can read the $k$ used from $b_{n+1}$ and undo the subtraction accordingly.
This shows
$$(n+1)^{n+1}\ge (n+2)^n. $$
To show that we can replace $\ge$ by $>$ (provided $n>0$) it suffices to exhibit a single $(a_1,\ldots,a_{n})$ that allows to construct at least two different $b_i$, i.e., where at least two choices of $k$ are possible. Indeed, for $(a_1,\ldots,a_n)=(0,\ldots,0)$ we might choose $k=1$ or $k=n+1$.
A: Let $a_1+\cdots+a_{n+1}=(n+1)^2$. Then multiplication of them gets maximum value when $a_1=\cdots=a_n=(n+1)$. The other combination is $a_1=\cdots=a_n=n+2$ and $a_{n+1}=1$.
A: Consider $(n+1)=a$ so our question becomes $a^a>(a+1)^{a-1}$ lhs=$a^a$ and  rhs after simplifying using binomial theorem and taking $a^a$ common can be written  as $a^a(\frac{1}{a}+\frac{a-1}{a^2}+...+\frac{1}{a^a})$ .now we can clearly see that the series in bracket is decreasing so its mod is less than or equal to $1$. Thus we its proved that lhs$>$rhs.
A: That can be written as:
$$ \left(\frac{n+2}{n+1}\right)^{n-1}\cdot\left(\frac{n+2}{n^2+2n+1}\right) < 1 $$
that follows from the AM-GM inequality, provided that $n\geq 1$:
$$ \left(\frac{n+2}{n+1}\right)^{n-1}\cdot\left(\frac{n+2}{n^2+2n+1}\right) \leq \left(\frac{n(n+2)}{(n+1)^2}\right)^n < 1.$$
A: (Not a combinatorial proof but a short one)
For $n=1$ we verify the inequality directly.
The function $x/\ln x$ is increasing from $x=e$ onwards. We need to prove that
$$(n+1)\ln(n+1)>n\ln(n+2),$$
which, for $n\geq 2,$ follows from
$$(n+1)(n+1)>n(n+2)\hbox{ and }\frac{\ln(n+2)}{n+2}<\frac{\ln(n+1)}{n+1}.$$
