Asymptotic to a sequence of algebraic numbers. Let $f(n)$ be the largest real solution of
$$x^n - x^{n-1} = 1 $$
As $n$ grows to positive infinity we get the asymptotic :
$$ f(n) = 1 + \frac{\ln(n)}{n} + \frac{\exp(2)}{n^2} + ...$$
Where the value $\exp(2)$ is optimal !
( and $...$ means smaller term(s) )
Notice $f(2)$ is the golden mean.
How to show this asymptotic  ?

Edit
Corrected the formula.
 A: Let $f(n)=1+\epsilon(n)$.  Then, $f^n(n)-f^{n-1}(n)=1$ becomes
$$(n-1)\log(1+\epsilon(n))+\log(\epsilon(n))=0$$
As $n\to \infty$, $\epsilon(n)\to 0$.  Hence, we have
$$(n-1)\epsilon(n)+O(n\epsilon^2(n))+\log(\epsilon(n))=0 \tag 1$$
We can write $(1)$ equivalently as
$$(n-1)\epsilon(n)e^{(n-1)\epsilon(n)}=(n-1)e^{O(n\epsilon^2(n))}\tag 2$$
which using Lambert's W function is given by
$$\epsilon(n)=\frac{1}{n-1}W\left((n-1)e^{O(n\epsilon^2(n))}\right)\tag 3$$
Using the first term in the large argument asymptotic expansion of $W$ yields
$$\begin{align}
\epsilon(n)&\sim \frac{1}{n-1}\log((n-1)e^{O(n\epsilon^2(n))})\\\\
&\sim\frac{\log(n-1)}{n-1}\\\\
&\sim\frac{\log(n)}{n}
\end{align}$$
Hence, we find that the first two terms in the expansion of $f(n)$ for large $n$ is given by 

$$\bbox[5px,border:2px solid #C0A000]{f(n)\sim 1+\frac{\log(n)}{n}}$$

And we are done!
A: A rough outline.
The first few terms in the asymptotic are
$$
f(n) = 1 + \frac{W(n)}{n} + \frac{W(n)^2}{2n^2} + \cdots,
$$
where $W$ is the Lambert-W function, so the stated asymptotic is incorrect.
First show that, with $x = 1 + \frac{W(n)}{n} + \frac{z}{n}$, where $z = O(1)$, we have
$$
x^n - x^{n-1} - 1 \to e^z - 1
$$
uniformly as $n \to \infty$. Conclude by Hurwitz's theorem that
$$
f(n) = 1 + \frac{W(n)}{n} + \frac{\epsilon_n}{n}
$$
with $\epsilon_n \to 0$ as $n \to \infty$. Substitute this into the equation
$$
f(n)^n - f(n)^{n-1}-1 = 0
$$
and apply asymptotic simplifications to conclude that $\epsilon_n \sim W(n)^2/(2n)$ as $n \to \infty$.
