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 $$u_n=f_{n+1}^{-1}\Big(f_{n+1}(u_{n})\Big)\geq f_{n+1}^{-1}\Big(f_{n+1}(u_{n+1})\Big)=u_{n+1}.$$
$$\color{red}{*}\color{blue}{: \begin{matrix} f_{n+1}(u_n) &=& u_n^{n+1}-u_n-(n+1) \\ &=&u_n u_n^n-u_n-(n+1) \\ &=&u_n(u_n+n)-u_n-(n+1)\\ &=&u_n^2+nu_n-u_n-n-1 \end{matrix}}$$
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 $\blacksquare$