If $ a_{n}\to+\infty $ then $(1+\frac{1}{a_{n}})^{a_{n}}\to e$ I already am given a solution to this that I don't quite understand. Here is how it goes:

If $n_{k}$ is any increasing sequence of positive integers then
$\lim_{k\rightarrow \infty } \ (1+\frac{1}{n_{k}+1})^{n_{k}}= \lim_{k\rightarrow \infty } \ (1+\frac{1}{n_{k}})^{n_{k}+1}=e$      [1]
The result follows from the following inequality using the Squeeze theorem and [1]:
$(1+\frac{1}{\left \lfloor a_{n} \right \rfloor+1})^{\left \lfloor a_{n} \right \rfloor}\leq (1+\frac{1}{a_{n}})^{a_{n}}\leq (1+\frac{1}{\left \lfloor a_{n} \right \rfloor})^{\left \lfloor a_{n} \right \rfloor+1}$

I have two questions:

*

*I don't know how to get that $\lim \ (1+\frac{1}{n_{k}+1})^{n_{k}}= \lim \ (1+\frac{1}{n_{k}})^{n_{k}+1}$


*A sequence $a_{n}$ that goes $1,10,1,100,1,1000,...$ diverges to $+\infty$ but isn't increasing so I think we shouldn't be able to use [1]. Is this proof erroneous?
Thanks in advance.
EDIT:
Thank you for all your insightful answers. You got me wondering now what if I had had:
$\lim \ (\frac{1}{n_{k}+1})^{n_{k}}= \lim \ (\frac{1}{n_{k}})^{n_{k}+1}$
The suggested method for 1. wouldn't have worked because we get the indeterminate form $0.\infty $ so how could one do then without the squeeze theorem?
 A: Hint : $$(1+\frac{1}{n_k+1})^{n_k+1}$$ and $$(1+\frac{1}{n_k})^{n_k}$$
both tend to $e$, when $n_k$ tends to $\infty$. Furthermore , $1+\frac{1}{n_k}$ and $1+\frac{1}{n_k+1}$ both tend to $1$, when $n_k$ tends to $\infty$.
Take it from here.
A: For question (1):  Observe that 
$$ \lim_{n\to\infty}\left(1+\frac{1}{n_{k}}\right)^{n_{k}+1}= \lim_{n\to\infty}\left(1+\frac{1}{n_{k}}\right)^{n_{k}}\cdot\lim_{n\to\infty}\left(1+\frac{1}{n_{k}}\right)=e\cdot1=e,$$
and 
$$ \lim_{n\to\infty}\left(1+\frac{1}{n_{k}+1}\right)^{n_{k}}=\frac{\lim_{n\to\infty}\left(1+\frac{1}{n_{k}+1}\right)^{n_{k}+1}}{\lim_{n\to\infty}\left(1+\frac{1}{n_{k}+1}\right)}=\frac{e}{1}=e. $$
Hence, the two limits are equal.
For question (2): We say that a sequence $\{a_n\}$ of real numbers diverges to $\infty$ if for any $M>0$ there is some $N\in\mathbb{N}$ such that for all $n\geq N$ we have $a_n>M$.  Does your example satisfy this criteria? 
In response to your edit:  Observe that 
$$\left(\frac{1}{{n_k}}\right)^{n_k}> \left(\frac{1}{{n_k}+1}\right)^{n_k}>0\implies\lim_{n\to\infty}\left(\frac{1}{{n_k}+1}\right)^{n_k}=0,$$
and 
$$\left(\frac{1}{{n_k}}\right)^{n_k}>\left(\frac{1}{{n_k}}\right)^{n_k+1}>0\implies\lim_{n\to\infty}\left(\frac{1}{{n_k}}\right)^{n_k+1}=0,$$
both by the squeeze theorem.  
A: *

*The limit of $(1+1/n_k)$ is $1$, so by multiplying and dividing by it, doesn't change the limit, thus the result derives from the well known $(1+1/n)^n \to e$ and a variable change!

*It doesn't diverge to infinity, its limit is undefined.
