Asymptotic to a sequence of algebraic numbers.

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!