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.

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 $(3)$

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

