The root of x^n + x = 1

Existence and uniqueness. $f_n$ is continuous and strictly increasing on $[0,1]$, with $f_n(0)=-1<0<1=f_n(1)$. By the intermediate value theorem and strict monotonicity there is exactly one root $x_n\in(0,1)$.

Monotonicity. Since $x_n^n=1-x_n$,
$$f_{n+1}(x_n)=x_n^{n+1}+x_n-1=x_n^{n+1}-x_n^n=x_n^n(x_n-1)<0=f_{n+1}(x_{n+1}).$$
As $f_{n+1}$ is increasing, $x_n<x_{n+1}$.

Limit. The sequence is increasing and bounded by $1$, so it converges to some $L\le1$. Suppose $L<1$. Then $x_n\le L$ for all $n$, so $x_n^n\le L^n\to0$, and $x_n=1-x_n^n\to1$. This contradicts $x_n\le L<1$. Hence $L=1$.

Numerics. $x_5=0.7549$, $x_{10}=0.8351$, $x_{100}=0.9666$, $x_{1000}=0.99476$.

Remark. One can show that $1-x_n\sim\frac{\ln n}n$, but the ratio approaches $1$ very slowly (about $0.76$ for $n=1000$). Such implicitly defined sequences are typical in analysis exams: you never get a formula, only properties.