Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.

How can I show this equation has no solutions for $n>5$.

Thanks.
 A: For $n \geq 6$ we have
$$\frac{(n+3)^n}{(n+2)^n} = \left( 1+\frac{1}{n+2} \right)^n = \left( 1+\frac{1}{n+2} \right)^{n+2} \cdot \left(\frac{n+2}{n+3}\right)^2$$
Both factors are monotonely increasing, hence this is at least its value for $n=6$, which is $\left(\frac{9}{8}\right)^6 \approx 2.072\dotsc > 2$.
Since $k \mapsto \frac{k+1}{k}$ is monotonely decreasing, we obtain
$$\frac{(k+1)^n}{k^n} \geq \frac{(n+3)^n}{(n+2)^n} > 2$$
for all $k = 1, \dotsc, n+2$, if $n \geq 6$. We can write this as
$$(k+1)^n-k^n > k^n$$
for all $k = 1, \dotsc, n+2$, if $n \geq 6$. This also holds for $k=0$.
In particular we have $$(n+3)^n = \sum_{k=0}^{n+2} ((k+1)^n-k^n) > \sum_{k=0}^{n+2} k^n = 1^n + 2^n + 3^n + \dotsb + (n+2)^n$$
for $n\geq 6$, an even stronger result.
A: Set $f(x)=\left(1-\frac{1}{x+3}\right)^x=\left(\frac{x+2}{x+3}\right)^x$ where $x\ge 6$. We have
$$f'(x)=\left(\frac{x+2}{x+3}\right)^x\left(\ln\left(\frac{x+2}{x+3}\right)+\frac{x}{x^2+5x+6}\right)$$
 $f'(x)<0$  for $x\ge 6$. Hence $f(x)\le f(6)=\left(\frac 89\right)^6<\frac 12$, i.e
$$\left(1-\frac{1}{x+3}\right)^x< \frac 12\tag 1$$
 Note
$$\sum\limits_{k=1}^{n}{{{(k+2)}^{n}}}={{(n+3)}^{n}}$$
thus
$$\sum\limits_{k=1}^{n}{{{\left( \frac{k+2}{n+3} \right)}^{n}}}=1$$
in other words
$$\sum\limits_{k=1}^{n}{{{\left( 1-\frac{n+1-k}{n+3} \right)}^{n}}}=1$$
or
$$\sum\limits_{k=1}^{n}{{{\left( 1-\frac{k}{n+3} \right)}^{n}}}=1$$
Set
$$I=\sum\limits_{k=1}^{n}{{{\left( 1-\frac{k}{n+3} \right)}^{n}}}\tag 2$$
By application of Bernoulli's inequality, we have
$$\left( 1-\frac{k}{n+3} \right)\le {{\left( 1-\frac{1}{n+3} \right)}^{k}}$$
therefore as $n\ge 6$
$${{\left( 1-\frac{k}{n+3} \right)}^{n}}\le {{\left( 1-\frac{1}{n+3} \right)}^{nk}}< \left(\frac12 \right)^k\tag 3$$
$(2)$ and $(3)$

$$\color{red}{I=\sum\limits_{k=1}^{n}{{{\left( 1-\frac{k}{n+3}
 \right)}^{n}}}<\sum_{k=1}^{n}\left(\frac12 \right)^k=1}$$

A: Let $n\ge5$ and suppose $$ \sum_{k=3}^{n+2}k^n<(n+3)^n.$$Then $$ \sum_{k=3}^{n+3} k^n <2(n+3)^n,$$ and we deduce \begin{align} \sum_{k=3}^{n+3} k^{n+1}<2(n+3)^{n+1}&<(n+4)^{n+1} \\ \left(1+\frac1{n+3}\right)^{n+1}\ge\left(\frac98\right)^6&>2, \end{align} where he have used the increasingness of $f(x)=\left(1+\frac1{x+3}\right)^{x+1}$ for $x>-3$: $f'(x)>0$ follows from $$ (x^2+6x+12)\log\left(1+\frac1{x+3}\right)>(x+3)^2((x+3)^{-1}-(x+3)^{-2})=x+2>x+1.$$
By induction, LHS < RHS for all $n\ge5$, and as you noted, it also holds for $n=4$.
A: A direct proof for $n\ge7$:
One can very easily show by induction $\sum_{k=1}^{n-1} k^n<n^n$. This implies \begin{align} \sum_{k=3}^{n+2}\left(\frac{k}{n}\right)^n &=1+\left(1+\frac1n\right)^n+\left(1+\frac2n\right)^n+\sum_{k=3}^{n-1} \left(\frac{k}{n}\right)^n \\ &< 2+e+e^2<\left(\frac{10}{7}\right)^7\le\left(1+\frac3n\right)^n.\end{align}
