How can I prove $\lim \limits_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e$ without involving function limit? If I already know that
$$\lim \limits_{n \to \infty} a_n=+\infty$$
Then how can I prove
$$\lim \limits_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e$$ 
without involving function limit?
This question comes because you may find some books on calculus or analysis (maybe they are badly written) require you to prove something like
$$\lim \limits_{n \to \infty}\left(1+\frac{2}{n}\right)^{n}=e^2$$
or something more complex even before they formally introduce the definition of limit of a function, They are hard to prove because you can't simply take something like $\frac{n}{2}$ as a subsequence of $n$. 
The definition of limit of a function (at infinity) here mean:

For a real function $f$ which is well-defined on $[a, +\infty)$, if for any $\epsilon >0$, there is a positive number $M \geq a$ such that when $x>M$ we can say $|f(x)-A|<\epsilon$, then
  $$\lim \limits_{x \to \infty}f(x)=A.$$

While the definition of limit of a sequence here mean:

For a sequence $\{a_n\}$, if for any $\epsilon>0$, there is a positive integer $N$ such that when $n>N$ we can say $|a_n-A|<\epsilon$, then $$\lim \limits_{n \to \infty}a_n=A.$$

I know sequence is a "special" kind of function whose domain is $\mathbb{N}$ and thus sequence limit is but a special case of function limit. Here I say avoid involving the idea of function limit means not to use the idea above but only to prove it by the "special case" below. After all, $(1+\frac{1}{a_n})^{a_n}$ is still a "special" function - a sequence.
p.s. $e$ is defined by
$$\lim \limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e.$$
My try so far:
Since $$\lim \limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e,$$ for every $\epsilon>0$, there is $N \in \mathbb{N}$ s.t. for all $n>N$,$$|\left(1+\frac{1}{n}\right)^n-e|<\epsilon.$$
Meanwhile, since $$\lim \limits_{n \to \infty} a_n=+\infty,$$ for $N' \in \mathbb{N}$ and $N'>N$, there is $N'' \in \mathbb{N}$ s.t. for all $n>N''$, $a_n>N'>N.$
However, if $a_n$ become bigger then $1+\frac{1}{a_n}$ will be smaller, and vise versa, so I don't know how to deal with $\left(1+\frac{1}{a_n}\right)^{a_n}.$
 A: For $$\displaystyle\lim_{n\to\infty }\left(1+\frac{k}{n}\right)^n=e^k$$ with $k\in\mathbb N$, do it by induction. The case $k=1$ is the definition. For $k>1$, you have that 
$$\left(1+\frac{k+1}{n}\right)^n=\left(\frac{n+k+1}{n}\right)^n=\left(1+\frac{k}{n}\right)^n\left(1+\frac{1}{n+k}\right)^n=\underbrace{\left(1+\frac{k}{n}\right)^n}_{\to e^k\ (hyp\ induction)}\underbrace{\left(1+\frac{1}{n+k}\right)^{n+k}}_{\to e}\underbrace{\left(1+\frac{1}{n+k}\right)^{-k}}_{\to 1}\underset{n\to\infty }{\longrightarrow }e^{k+1}$$
A: Any subsequence of a Cauchy sequence is a Cauchy sequence with the same limit point, hence
$$ \lim_{n\to +\infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e $$
as soon as $\{a_n\}_{n\in\mathbb{N}}$ is a diverging sequence of natural numbers. The very last assumption can be dropped by noticing that $\frac{\log(1+x)}{x}$ is a positive, decreasing and convex function on $[0,1]$, hence if $a_m\in\mathbb{R}$ is between $n$ and $n+1$,
$$\left(1+\frac{1}{a_m}\right)^{a_m}$$
is between $\left(1+\frac{1}{n}\right)^n$ and $\left(1+\frac{1}{n+1}\right)^{n+1}$, so the same conclusion as above follows by squeezing.
A: You can also evaluate the limit with l'Hospitals rule. The Limit $\big(1+\frac{1}{a_n}\big)^{a_n}$ with $\lim_{n\to\infty} a_n=+\infty$ is a limit of the type $1^\infty$. As a replacement for $a_n\to\infty$ I would like to use $k\to\infty$. With this you get:
$$\lim_{k\to\infty} \big(1+\frac{1}{k}\big)^{k} = \lim_{k\to\infty} \mathrm{e}^{\ln\bigg(\big(1+\frac{1}{k}\big)^{k}\bigg)} = \lim_{k\to\infty} \mathrm{e}^{k\ln\big(1+\frac{1}{k}\big)}$$
Since the exponential function is a continous function, you can apply a limit theorem on the last expression.
$$\lim_{k\to\infty} \big(1+\frac{1}{k}\big)^{k} = \mathrm{e}^{\lim_{k\to\infty} k \ln\big(1+\frac{1}{k}\big)}$$
Inside of the exponential function you get a limit of the type $\infty\cdot 0$ which can be transformed to a limit of the type $\frac{0}{0}$ by doing the following:
$$\lim_{k\to\infty} \big(1+\frac{1}{k}\big)^{k} = \mathrm{e}^{\lim_{k\to\infty} \frac{\ln\big(1+\frac{1}{k}\big)}{\frac{1}{k}}}$$
You can apply l'Hospitals rule on the inner limit and you will get $1$ as result. So the limit will be $\mathrm{e}^1$.
