<s>[INST] The monotony of an implicit sequence Context of the exercice:
Let us define $(f_n)_n$ for any $n\geq 3$,  $\left\{\begin{matrix}
f_n:]1,2]&\rightarrow & \mathbb{R}  \\ 
 x &\mapsto & f_n(x):=x^n-x-n
\end{matrix}\right.$
It's easy to verify that $f_n$ is a bijection between $]1,2]$ and $]-n,f_n(2)]$ (since $f_n$ has a strictly positive derivative and $f_n(2)>0$).
Hence there exist a unique $u_n $ in $]1,2]$ such that $f_n(u_n)=0$.
Conclusion:
The sequence $$\left\{\begin{matrix}
u_n &\in & ]1,2]\\ 
 u_n^n &=& u_n+n
\end{matrix}\right.$$
is well defined.

Now the goal is to study the monotony of $(u_n)_n$.
I'm almost sure that $(u_n)_n$ is a decreasing sequence, I tried to prove it but I could not.
The idea is to show that $\color{red}{^{*}}$: $\color{blue}{f_{n+1}(u_n)>0\text{(why!?)}}$ and use the fact that $0=f_{n+1}(u_{n+1})$ and  that $f_{n+1}^{-1}$ is an increasing function to get $$u_n=f_{n+1}^{-1}\Big(f_{n+1}(u_{n})\Big)\geq f_{n+1}^{-1}\Big(f_{n+1}(u_{n+1})\Big)=u_{n+1}.$$
$$\color{red}{*}\color{blue}{: \begin{matrix}
f_{n+1}(u_n) &=& u_n^{n+1}-u_n-(n+1) \\ 
 &=&u_n u_n^n-u_n-(n+1) \\ 
 &=&u_n(u_n+n)-u_n-(n+1)\\
 &=&u_n^2+nu_n-u_n-n-1
\end{matrix}}$$
 [/INST] First of all, note that $u_n>1+\frac{1}{n}$ since
$$
{f_n(1+\frac{1}{n})=(1+\frac{1}{n})^n-1-n-\frac{1}{n}<e-1-n<e-4<0
\\\text{and $f_n(x)$ is increasing}
}
$$
To prove why $u_n^2+(n-1)u_n-(n+1)$ is always positive for $n\ge 3$ and $u_n>1+\frac{1}{n}$, note that the roots of $x^2+(n-1)x-(n+1)$ are
$$
{r_1=\frac{1-n-\sqrt {n^2+2n+5}}{2}<0<u_n\\
r_2=\frac{1-n+\sqrt {n^2+2n+5}}{2}
\\=\frac{2n+2}{\sqrt {n^2+2n+5}+n-1}
\\<\frac{2n+2}{n+1+n-1}=1+\frac{1}{n}<u_n
}
$$
hence, both of the roots of $x^2+(n-1)x-(n+1)$ are less than $u_n$ and since the coefficient of $x^2$ is positive, then $u_n^2+(n-1)u_n-(n+1)>0$ and the proof is complete $\blacksquare$
</s>