A conjecture related to an equation with floor functions 
Question : Is the following conjecture true?
Conjecture : Letting
  $
g_{n,k}(x)
=
\left\lfloor{\,\sqrt[n]{\,x\,}\,{\left\lfloor{
\,\sqrt[n]{\,x\,}\,\left\lfloor\cdots\left\lfloor
\,\sqrt[n]{\,x\,}\,\right\rfloor\cdots\right\rfloor}\right\rfloor}}\right\rfloor
$
  with $k$ $\,\sqrt[n]{\,x\,}\,$'s, an equation
  \begin{align}
g_{n,n}(x) + \sum_{i = 1}^{n - 1}2^{i - 1}g_{n,n - i}\left(x\right)
+ 2^{n - 1} - 1 = x\qquad(\star)\end{align}
holds if and only if $x=l^n-1\ (l\ge 2\in\mathbb N)$. 
Here, $\left\lfloor\,x\,\right\rfloor$ is the largest integer not greater than $x$.

Example : The $n=4$ case of $(\star)$ is the following : 
The conjecture above states that the following equation holds if and only if $x=l^4-1\ (l\ge 2\in\mathbb N)$ such as $x=15, 80, 255, 624,\cdots$.
$$\lfloor\sqrt[4]{x}\lfloor\sqrt[4]{x}\lfloor\sqrt[4]{x}\lfloor\sqrt[4]{x}\rfloor\rfloor\rfloor\rfloor+\lfloor\sqrt[4]{x}\lfloor\sqrt[4]{x}\lfloor\sqrt[4]{x}\rfloor\rfloor\rfloor+2\lfloor\sqrt[4]{x}\lfloor\sqrt[4]{x}\rfloor\rfloor+4\lfloor\sqrt[4]{x}\rfloor+7=x.$$
Motivation : I've known the following question :
"Find every $x\gt0\in\mathbb R$ such that $\lfloor\sqrt{x}\lfloor\sqrt{x}\rfloor\rfloor+\lfloor\sqrt{x}\rfloor+1=x$."
The answer is $x=(n+1)^2-1\ (n\in\mathbb N)$. This is in the form of $A^2-1$. Then, I got interested in finding a similar equation whose solutions are in the form of $A^n-1$. This conjecture is true for $n\le 6, x\le 100000$ by using computer. However, I don't have any good idea for proving this conjecture. Can anyone help?
 A: I have what might be a first step towards a solution. Throughout, I assume $n\ge2$ and $x\ge2$.
First, let
$$
G_n(x)=g_{n,n}(x)+g_{n,n-1}(x)+2g_{n,n-2}(x)+\cdots+2^{n-1}g_{n,0}(x)
$$
where $g_{n,0}(x)=1$. So, we wish to solve $G_n(x)=x+1$.
If $x=l^n$, we get $g_{n,k}(l^n)=l^k$, and so $G_n(l^n)=l^n+(l^n-2)/(l-2)>l^n+1$, which rules out that case.
Assume $l-1<u=\sqrt[n]{x}<l$ for some integer $l$. We write $g_{n,1}(x)=u-\epsilon_1$, and then by induction get the expression
$$
g_{n,k}(x)=\lfloor u\cdot g_{n,k-1}(x)\rfloor
=u^k-u^{k-1}\epsilon_1-\cdots-u\epsilon_{k-1}-\epsilon_k
$$
where $\epsilon_k\in[0,1)$ for $k=1,\ldots,n$. These, we plug into the expression for $G_n(x)$, use $\delta_k=1-\epsilon_k>0$ for simplification, and obtain
$$
G_n(x)=u^n+u^{n-1}\delta_1+u^{n-2}(\delta_1+\delta_2)+\cdots
+(2^{n-2}\delta_1+\cdots+\delta_{n-1}+\delta_n)
$$
where we immediately see that $G_n(x)>x$, hence, $G_n(x)\ge x+1$.
In order to have $G_n(x)=x+1$, the $\delta_k$ need to be strongly bounded. However, I haven't quite figured out how to use this to limit $x$ to $x=l^n-1$.
In particular, $\delta_1<1/u^{n-1}$ is required, which implies that
$l=u+\delta_1$ is an integer with $l<u(1+1/x)$. This gives
$$
l^n<u^n\left(1+\frac{1}{x}\right)^n<x+n
$$
which at least ensures $l^n-n<x<l^n$. I suspect better use of the bounds on the $\delta_k$ may help, but would like to see a less technical proof.
If we plug in $x=l^n-1$, we get $u\approx l-l/nx$. This should be enough to ensure $\delta_k\approx u^k/nx$ which I think should be sufficient to ensure $G_n(x)=u^n+1=x+1$ if $u>2$. However, I haven't had the time to go into the details of this.

I think I see how the rest goes, although what I provide here might not be very complete.
So, let's $\delta=\delta_1=l-u<1/u^{n-1}$, so we have $u=\sqrt[n]{x}=l-\delta$. Let $g_k=g_{n,k}(x)$, so we have $g_1=l-1$. Now, we prove by induction that
$g_k=l^k-l^{k-1}-\ldots-l-1$. Assuming this for $g_{k-1}$ and using $g_k=ug_{k-1}-1+\delta_k$, we have
$$
g_k=(l-\delta)g_{k-1}-1+\delta_k
=(l^k-l^{k-1}-\ldots-l-1)-\delta g_{k-1}+\delta_k
$$
where $\delta_k=\delta g_{k-1}<\delta u^{k-1}<1$, giving the desired expression for $g_k$. Note how $\delta u^{n-1}<1$ is required to make this work.
Now that $g_k=l^k-l^{k-1}-\ldots-1$, plugging these into the expression for $G_n$, we get $G_n=l^n$. To show this, we use induction to obtain $g_k+2g_{k-1}+\cdots+2^kg_0=l^k+\cdots+1$: it holds for $g_0=1$, and if it holds for $k-1$ we get
$$
g_k+2g_{k-1}+\cdots+2^kg_0=g_k+2\cdot(l^{k-1}+\cdots+1)=l^k+\cdots+1.
$$
This makes $G_n=g_n+(g_{n-1}+2g_{n-2}+\cdots+2^{n-1}g_0)=l^n$.
Thus, having $G_n=x+1$ requires $x=l^n-1$; the converse should also follow.
