I'm having trouble doing calculations with the little-o notation. Let's say that the following asymptotic formula holds for $n\to\infty$: $$ B_n = \frac{1}{\sqrt{r+1}} \exp\left( n \left(r+\frac{1}{r}-1\right)-1 \right)(1+o(1)) $$ where $r$ is the (unique and positive) solution to $re^r=n$. In the following you can assume that you can use the same $r$ for $B_n$, $B_{n+1}$ and $B_{n+2}$ (you can show that with some more asymptotics).
Somehow you can then prove the identity $$\frac{B_{n+1}}{B_n} -1 = \frac{n}{r} (1+o(1))$$ and $$ \frac{B_{n+2}}{B_n} - \left( \frac{B_{n+1}}{B_n} \right)^2 -1 = \frac{n}{r^2} (1+o(1)). $$ How do I go about proving these formulas?
I attempted to derive the first identity by just plugging in the asymptotic formula and got stuck: $$ \begin{align*} \frac{B_{n+1}}{B_n} - 1 &= \frac{ \frac{1}{\sqrt{r+1}} \exp\left((n+1)\left(r+\frac{1}{r}-1\right)-1\right) (1+o(1)) }{ \frac{1}{\sqrt{r+1}} \exp\left(n\left(r+\frac{1}{r}-1\right)-1\right) (1+o(1)) } -1 \\\\ &= \exp\left(r+\frac{1}{r}-1\right)(1+o(1)) - 1 \\\\ &= e^r e^{1/r} e^{-1} (1+o(1)) -1 \\\\ &= \frac{n}{r} e^{-1} (1+o(1)) -1 . \end{align*} $$
In the last equality I used $e^r = \frac{n}{r}$ and that $e^{1/r}=1+o(1)$ (at least I think that's true b/c $n\to\infty$ implies $r\to\infty$ and $e^{1/r}\xrightarrow{r\to\infty} 1$ and afaik that's precisely what $1+o(1)$ means for $r\to\infty$). But now there's still the term $e^{-1}$ left, of which I can't say it's $1+o(1)$, b/c $e^{-1}(1+o(1)) = e^{-1} + o(1)$. So somewhere I made a mistake but I don't know where...
(The $B_n$ are Bell numbers btw.)