On the behaviour of $\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n$ I have to find the limit : (let $k\in \mathbb{R}$)

$$\lim_{n\to \infty}n^k \left(\Big(1+\frac{1}{n+1}\Big)^{n+1}-\Big(1+\frac{1}{n}\Big)^n \right)=?$$

My Try :
$$\lim_{n\to \infty}\frac{n^k}{\Big(1+\frac{1}{n}\Big)^n} \left(\frac{\Big(1+\frac{1}{n+1}\Big)^{n+1}}{\Big(1+\frac{1}{n}\Big)^n}-1\right)$$
we know that :
$$\frac{\Big(1+\frac{1}{n+1}\Big)^{n+1}}{\Big(1+\frac{1}{n}\Big)^n}>1$$
now what do i do ?
 A: $$\left(1+\frac{1}{n}\right)^n = \exp\left[n\log\left(1+\frac{1}{n}\right)\right]=e-\frac{e}{2n}+O\left(\frac{1}{n^2}\right) $$
hence
$$ \left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n = \frac{e}{2n^2}+O\left(\frac{1}{n^3}\right) $$
and for a fixed $k\in\mathbb{R}$
$$ \lim_{n\to +\infty}n^k\left[\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right]$$
is non-trivial only if $k=2$. Otherwise, it is either $0$ or $+\infty$.
A: $$\lim_{n\to \infty}n^k \left((1+\frac{1}{n+1})^{n+1}-(1+\frac{1}{n})^n \right)=
\lim_{n\to \infty}n^k \left(\frac{e}{2n^2}+O((\frac{1}{n^3})) \right)$$
for n<2 limit is 0, for n=2 limit is e/2, for n>2 limit is infinity
A: Using only the Binomial Theorem and Bernoulli's Inequality:
$$
\begin{align}
\hspace{-1cm}\left(1+\frac1{n+1}\right)^{n+1}\!\!-\left(1+\frac1n\right)^n
&=\left(\frac{n+2}{n+1}\right)^{n+1}-\left(\frac{n+1}n\right)^n\tag{1a}\\
&=\color{#C00}{\left(\frac{n+1}n-\frac1{(n+1)n}\right)^{n+1}}-\color{#090}{\left(\frac{n+1}n\right)^n}\tag{1b}\\
&=\color{#C00}{\left(\frac{n+1}n\right)^{n+1}-(n+1)\left(\frac{n+1}n\right)^n\frac1{(n+1)n}}\\
&\color{#C00}{{}+\frac{(n+1)n}2\left(\frac{n+1}n\right)^{n-1}\left(\frac1{(n+1)n}\right)^2+O\!\left(\frac1{n^3}\right)}\\
&-\color{#090}{\left(\frac{n+1}n\right)^n}\tag{1c}\\
&=\frac{(n+1)n}2\left(\frac{n+1}n\right)^{n-1}\left(\frac1{(n+1)n}\right)^2+O\!\left(\frac1{n^3}\right)\tag{1d}\\
&=\frac1{2(n+1)n}\left(\frac{n+1}n\right)^{n-1}+O\!\left(\frac1{n^3}\right)\tag{1e}\\[3pt]
&=\frac{e}{2n^2}+O\!\left(\frac1{n^3}\right)\tag{1f}
\end{align}
$$
Explanation:
$\text{(1a)}$: combine terms
$\text{(1b)}$: add and subtract $\frac1{(n+1)n}$
$\text{(1c)}$: expand the first three terms of the Binomial expansion
$\text{(1d)}$: the sum of the first two terms of the Binomial expansion
$\phantom{\text{(1d):}}$ equals $\left(\frac{n+1}n\right)^n$
$\text{(1e)}$: simplify
$\text{(1f)}$: $\frac1{(n+1)n}=\frac1{n^2}+O\!\left(\frac1{n^3}\right)$  and $\left(\frac{n+1}n\right)^{n-1}=e+O\!\left(\frac1n\right)$
$\phantom{\text{(1f):}}$ the latter equation uses this answer to show that
$\phantom{\text{(1f):}}$ $\left(1+\frac1n\right)^n\le e\le\left(1+\frac1n\right)^{n+1}$ with Bernoulli's Inequality
