Convergence of $\sqrt[k]{z+\sqrt[k]{z+\sqrt[k]{z+\cdots}}}$, where $z=(1+x)^k-(1+x)$ If one writes
$$1+x=\sqrt{(1+x)^2}=\sqrt{1+2x+x^2}=\sqrt{x+x^2+(1+x)}$$
then one has a recursive definition of the function $1+x$ which can be used to write $1+x$ as the infinite nested radical:
$$1+x=\sqrt{x+x^2+\sqrt{x+x^2+\sqrt{x+x^2+\sqrt{\cdot\cdot\cdot}}}}$$
But this definition relies on the fact that
$$
1+x=\sqrt{(1+x)^2}
$$
which is only true for $x \ge-1$. In general one could state that
$$
1+x=\sqrt[n]{(1+x)^n}=\sqrt[n]{(1+x)^n-(1+x)+\sqrt[n]{(1+x)^n-(1+x)+\sqrt[n]{\cdots}}}
$$
But the RHS does not converge to $1+x$ for most values of $x\in\mathbb{C}$. So, my question is, what is the actual closed form of the following function? For what values of $x\in\mathbb{C}$ does the following function converge?
$$
\sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{\cdots}}}
$$
If unclear the above nested radical can be defined by $\lim_{n\to\infty}a_n$ where
$$
a_1=\sqrt[k]{(1+x)^k-(1+x)},\quad
a_n=\sqrt[k]{(1+x)^k-(1+x)+a_{n-1}}.$$
 A: The expression 
$$
y=\sqrt[k]{(1+x)^k-(1+x)+\sqrt[k]{(1+x)^k-(1+x)+\cdots}}
$$
where $x\ge 0$, represents the limit of the recursive sequence
$$
a_1=\sqrt[k]{(1+x)^k-(1+x)}, \quad a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n},
\quad n\in\mathbb N.
$$
if such a limit exists.
Clearly, the sequence $\{a_n\}$ is increasing. (The fact $a_n\le a_{n+1}$ can be shown inductively.)
Next, we show inductively that $\{a_n\}$ is upper bounded by $1+x$. 
Clearly, 
$$
a_1=\sqrt[k]{(1+x)^k-(1+x)}\le\sqrt[k]{(1+x)^k}=1+x.
$$
Assume that $a_n\le 1+x$. Then
$$
a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n}\le
\sqrt[k]{(1+x)^k-(1+x)+(1+x)}=1+x. 
$$
Therefore, $\{a_n\}$ is increasing and upper bounded and thus it converges. Let $y=\lim a_n$.
If $x=0$, then observe that $a_n=0$, for all $n$, and hence $y=0$.
If $x>0$, then clearly $y>0$, and hence
$$
y \leftarrow a_{n+1}=\sqrt[k]{(1+x)^k-(1+x)+a_n}\to 
\sqrt[k]{(1+x)^k-(1+x)+y}
$$
and thus
$$
y^k-y=(1+x)^k-(1+x)
$$
Now the function $g(z)=z^k-z$ is strictly increasing, and hence one-to-one when $g'(z)=kz^{k-1}-1>0$ equivalently when $z>k^{-1/(k-1)}$. So if we show that $y>k^{-1/(k-1)}$, then we will shown that $y=1+x$.
We have
$$
a_1=\sqrt[n]{(1+x)^k-(1+x)}=\sqrt[n]{(1+kx+\cdots)-(1+x)} \\ \ge
\sqrt[n]{(k-1)x}\ge x^{1/k}
$$
then
$$
a_2=\sqrt[n]{(1+x)^k-(1+x)+a_1}\ge a^{1/k}_1\ge x^{1/{k^2}}
$$
and in general 
$$
a_n\ge x^{1/{k^n}}\to 1,
$$
and hence
$$
y=\lim a_n\ge 1>k^{-1/(k-1)},
$$
which implies that $a_n\to 1+x$.
Note. If $x\in [-1,0]$, then $a_n\to 0$.
