Positive real solution to $x^n=x+1$ I am interested in finding the positive real solution to $x^{n}=x+1$, where $n<0, n>1$. For example, if $n = 2$, then $x$ is the golden ratio: $\frac{1+\sqrt{5}}{2} \approx 1.618$. As $n$ approaches $\infty$, $y$, the solution to the equality, approaches $1$.
To find the solution, I fixed $y$. Then $$y^n=y+1$$ Solving for $n$ yields $$n = \log_{y}(y+1)$$
We now have a solution only in terms of $n$ and $y$. However, I would like to find $y$ as a function of $n$, not the other way around.
If this is not possible, as I suspect, I would like to find an approximation. For my attempt to find an approximation, I examined $$n = \log_{s}(s+1)$$ with $s$ being $1-\frac{1}{y_1}$. I graphed this and got something that asymptotically looked like a line, with a gap from $n = 0$ to $n = 1$, of the form $y_1=kn+1$, where $k \approx -1.4427$. This then means that $$\frac{1}{1-y}=kn+1 \rightarrow y \approx -\frac{1}{kn+1}+1$$ . This matches up with the graph. In the graph, $n$ is on the $x$-axis.
My questions: 
Is there a closed form for the inverse of $n = \log_y(y+1)$?
What is the exact value of $k$?
What is a better approximation, perhaps of the form $y=c_0+\frac{1}{c_1n+d_1}+ \frac{1}{c_2n^2+d_2} +...$?
Edit:
Letting $$t = 1-\frac{1}{y_1-1+\frac{1}{kn+1}}$$ I graphed $n=\log_t(t+1)$ This was asymptotically equal to $$y_1=kn+2$$
However, I'm not sure how to use this in order to get a better estimate.
Edit 2:
I appreciate the answers so far. However, using a simple sum of inverse powers will not work near $x = 0$. At $x = 0$, the approximations of the form $y=a_0+a_1x^{-1}+a_2x^{-2}+...$ will diverge to $-\infty$. In fact, successive approximations get worse for $-1<x<0$. On the other hand, $$y = -\frac{1}{kn+1}+1$$ approaches $0$ as $n$ approaches $0$.
 A: Let us consider the equivalent equation $x = (x+1)^{1/n}$; then, since we are interested in the behavior as $n \to \pm\infty$ (and since $\frac{1}{n}$ already appears in the equation), let us set $\epsilon := \frac{1}{n}$.
Now, consider the function $F(\epsilon, x) := x - (x+1)^\epsilon$.  We have that $F$ is differentiable in a neighborhood of $(0, 1)$; $F(0, 1) = 0$; and $\frac{\partial F}{\partial x}(0, 1) = 1 \ne 0$.  Therefore, by the Implicit Function Theorem, there is a differentiable function $f(\epsilon)$ on a neighborhood of $\epsilon = 0$ such that $f(0) = 1$ and $F(\epsilon, f(\epsilon)) = 0$.  This function $f(\epsilon)$ is exactly the solution to the equation that we were looking for as a function of $\epsilon$.
Furtherfore, $f'(0) = -\frac{\partial F/\partial \epsilon(0, 1)}{\partial F/\partial x(0, 1)} = \ln 2$.  It follows that $f(\epsilon) = 1 + \epsilon \ln 2 + o(\epsilon) = 1 + \frac{\ln 2}{n} + o(\frac{1}{n})$.
If we want to get more accurate approximations, then observe that since $F$ is $C^\infty$ near $(0, 1)$, then $f \in C^\infty$ in a neighborhood of 0 also, so Taylor series approximations will give the desired refinements.  The higher derivatives of $f$ at 0, in turn, can be found by a standard implicit differentiation of the equation $f(\epsilon) - (f(\epsilon) + 1)^\epsilon = 0$.
A: The root is,
for large $n$,
$1+\frac{\ln(2)}{n}+O(\frac1{n^2})
$.
Let
$f(x) = x^n-x-1$.
$f(0) = -1,
f(1) = -1,
f(2) = 2^n-3
\gt 0$
for
$n \ge 2$.
$f'(x)
=nx^{n-1} -1
\ge n-1$
for $x \ge 1$.
Therefore
only one real root
with $x > 1$.
Let
$x_0$ be that root.
Since
$(1+\frac1{n})^{n+1}
\gt e$,
$f(1+\frac1{n-1})
=(1+\frac1{n-1})^n-(1+\frac1{n-1})-1
\gt e-2-\frac1{n-1}
\gt 0$
for $n-1 \gt \frac1{.7}$
or $n \ge 3$.
Therefore
$1 \lt x_0
\lt 1+\frac1{n-1}
$.
Let's look for a root
of the form
$1+\frac{c}{n}
$.
$\begin{array}\\
f(1+\dfrac{c}{n})
&=(1+\dfrac{c}{n})^n-(1+\dfrac{c}{n})-1\\
&=e^{n\ln(1+c/n)}-2-\dfrac{c}{n}\\
&=e^{n(c/n-c^2/(2n^2)+O(1/n^3)))}-2-\dfrac{c}{n}\\
&=e^{c-c^2/(2n)+O(1/n^2)}-2-\dfrac{c}{n}\\
&=e^{c}e^{-c^2/(2n)+O(1/n^2)}-2-\dfrac{c}{n}\\
&=e^{c}(1-c^2/(2n)+O(1/n^2))-2-\dfrac{c}{n}\\
&=e^{c}-e^{c}(c^2/(2n)+O(1/n^2))-2-\dfrac{c}{n}\\
&=e^{c}-2-\frac{c}{n}(1+\frac{ce^{c}}{2})+O(1/n^2)\\
\end{array}
$
Therefore,
if $e^{c} = 2$,
or
$c = \ln(2)
$,
$f(1+\dfrac{c}{n})
=-\frac{\ln(2)}{n}(1+\ln(2))+O(\frac1{n^2})
$
and,
for any other $c$,
$f(1+\dfrac{c}{n})
=e^c-2 +O(\frac1{n})
$.
