I was not clear in my last post :
Let $0<x$ a real number then we have : $$x^{1-x+x^{1-x+x^{1-x+x^{1-x+\cdots}}}}=x$$
I take the logarithm of both side we get :
$$\ln(x)({1-x+x^{1-x+x^{1-x+x^{1-x+\cdots}}}})=\ln(x)$$
We can simplify into :
$${1-x+x^{1-x+x^{1-x+x^{1-x+\cdots}}}}=1$$
Wich is true recalling the initial equality .
Edit : I prove nothing but see the comment above .
I would learn if you have other proof ?
thanks in advance .
Firstly let us define the LHS expression. I will assume that we define it as follows $$x^{1-x+x^{1-x+x^{1-x+x^{1-x+\cdots}}}}=\lim_{n\to\infty}a_n$$ Where $a_n$ is the sequence defined by the recurrence relation $$a_1=x^{1-x}\qquad a_{n+1}=x^{1-x+a_n}$$ Now consider the subsequences given by $b_n=a_{2n}$ and $c_n=a_{2n-1}$. These sequences can be defined by the recurrence relations $$b_1=x^{1-x+x^{1-x}}\qquad b_{n+1}=f(b_n)$$ $$c_1=x^{1-x}\qquad c_{n+1}=f(c_n)$$ where the function $f:(0,1)\mapsto(0,1)$ is defined by $f(y)=x^{1-x+x^{1-x+y}}$. It is possible to then prove that $$y\lt x\implies y\lt f(y)\lt x$$ $$y\gt x\implies x\lt f(y)\lt y$$ Then, noting that $b_1\lt x$ and $c_1\gt x$, we can see that $b_n$ is a strictly increasing sequence and $c_n$ is a strictly decreasing sequence. Using these statements we can also show that $b_n\lt x$ and $c_n\gt x$ for all $n\in\mathbb{N}$. So, $b_n$ and $c_n$ converge by the monotone convergence theorem. We can then show that $b_n$, $c_n$ (and hence $a_n$) converge to $x$ as $n\to\infty$ by solving the equation $f(y)=y$ which has the unique solution $y=x$ for $0\lt x\lt1$. Thus we can finally say that $$x^{1-x+x^{1-x+x^{1-x+x^{1-x+\cdots}}}}=\lim_{n\to\infty}a_n=x$$