Calculate and prove the limit $\lim\limits_{n\to\infty }(1+\frac{1}{a_n})^{a_n}$ Given $a_n$ is an increasing monotone sequence of integers 
$\lim\limits_{n\to\infty }(1+\frac{1}{a_n})^{a_n}$ given $(a_n)$ is an increasing sequence of integers

NOTE - I noticed these 2 questions here (1) and here (2) but I believe that my question is a bit different because these questions answer only the case of $a_1 \geq 0$
according to the information we can understand that if $a_1 \geq 0$ then it is immediately solved as $\lim\limits_{n\to\infty }(1+\frac{1}{a_n})^{a_n} =e$ since then we can look at $a_n$ as in $a_n=n$ and then the limit would be just an identity $\lim\limits_{n\to\infty }(1+\frac{1}{n})^{n}=e$
otherwise what if $a_1 <0$? we will need to show that  there exists an $N \in \Bbb N$ such that $a_N \geq 0$  we can also understand that $a_{n+1}-a_n \geq 1$ then for ever $n$ we get $a_{n+1}-a_n = (a_{n+1}-a_n)+(a_n-a_{n+1})...+(a_2 - a_1) \geq 1+1+1...+1 =n$
I do not know how to continue from here.. I usually have more ideas and stuff I tried on my posts but I really cannot figure out what to do here.
thanks for any help and tips!
edit - Thanks to all the comments I tried and cannot figure out on how to actually prove it , I do realize what the limit is worth now and why but I am struggling to prove it as I stated before
EDIT (point of the edit is to solve it organized and in a better way according to information I have and from the comments) -
Organizing information
a.
$(a_n)$ is an increasing sequence therefore we get $(1)$ $a_{n+1} > a_n$ , $(2)$ thanks to nejimban an increasing sequence of integers must tend to $\infty$ $(3)$ an increasing sequence of integers therefore $a_{n+1}-a_n \geq 1$
b. $\lim\limits_{n\to\infty }(1+\frac{1}{n})^{n}=e$
c. Need to prove that there is an $N \in \Bbb N$ such that for every $n>N$ we get $a_n>0$
Solving - posted as an answer to the question
 A: Hopefully this is the right answer after all the amazing help in the comments and hopefully this can help someone if they struggled with this like I did.
For now I will accept my own answer as it seems logical to me but if someone finds something wrong with it please let me know
First we organize given information , conclusions and theorems we will use.

*

*Given $(a_n)$ is an increasing sequence of integers therefore $a_{n+1}-a_{n}\geq1$

*$a_{n+1} > a_n$

*theorem : an increasing sequence which is not bounded must tend to $\infty$

*$\lim\limits_{n\to\infty }(1+\frac{1}{n})^{n}=e$  is a known limit

*a subsequence of a convergent sequence also converges to the original series limit

*sequence $(b_n)$ is called a subsequence of $a_n$ if there is an increasing sequence $(n_k)$ of natural numbers that satisfies $a_{n_k}=b_k$ for every $k \in \Bbb N$
Solution:
We will first prove that $\lim\limits_{n\to\infty }a_n=\infty$ for the case $a_1<0$ (if $a_1 \geq 0$ there isn't much to prove).
let $a_1 = -k \in \Bbb Z$.
therefore $a_{n+t} \geq t+a_1=0$ (as podiki said) we are checking $a_1 <0$ so we picked a first element which is an integer so let $a_1 = -t$ . if $a_1 = -t$ then since the series is increasing after $t$ terms we must have a nonnegative number since the sequence goes up by at least one each term then we get $a_{n+t} \geq 0$.
now let $N \in \Bbb N$ such that for all $n>N$ we get $a_n >a_N \geq0$ and we found such an $n$  ( $a_{n+t} \geq 0$ ) according to point 2 and point 1 we get that $a_{n+t+1} > a_{n+t}$ therefore according to point 6 this is an increasing sequence of natural numbers. then we can say that $(1+ \frac{1}{a_{n+t}})^{a_{n+t}}$ is a subsequence of $(1+ \frac{1}{n})^n$ and since according to point 4 we know that $\lim\limits_{n\to\infty }(1+\frac{1}{n})^{n}=e$ then $\lim\limits_{n\to\infty }(1+ \frac{1}{a_{n+t}})^{a_{n+t}}=e$ as well, and since  $\lim\limits_{n\to\infty }(1+ \frac{1}{a_{n+t}})^{a_{n+t}}=e$ is a moved sequence of $a_n$ (sorry if it is not called "moved sequence" in english hope it is still understood) we get that $\lim\limits_{n\to\infty }(1+ \frac{1}{a_{n}})^{a_{n}}=e$
A: I think your proof can be written out a little more clearly, and with fewer steps, words and symbols. In particular, I will only use one "lemma" before moving to the main proof.

Lemma: For each $n\in\mathbb N$, $a_n \geq a_0 + n$.
Proof: Induction. For $n=0$, the proof is trivial. For general $n$, assume $a_n\geq a_0 + n$. Then, $a_{n+1} > a_n$ and $a_{n+1}$ is an integer, so $$a_{n+1}\geq a_n + 1 \geq (a_0 + n) + 1 = a_0 + (n+1)$$ which concludes the proof.


Main proof. Let $$b_n=\left(1+\frac1{a_{n'}}\right)^{a_{n'}}.$$
Let $\epsilon > 0$. Then, from the definition of limits and because we know that
$$\lim_{n\to\infty}\left(1+\frac1n\right)^n = e$$
we know that there exists some $M\in\mathbb N$ such that, for all $n\in \mathbb N$ such that $n\geq M$, we have $$\left|e-\left(1+\frac1n\right)^n\right|\leq\epsilon.$$
Let $M' = M - a_0$. Let $n'\in\mathbb N$ be such that $n' > M'$. Then, we have $$a_{n'} \geq a_0 + n' \geq a_0 + M' \geq a_0 + M-a_0 \geq M.$$
Therefore, we know that $a_n' \geq M$, which means that
$$|e-b_{n'}|=\left|e-\left(1+\frac1{a_{n'}}\right)^{a_{n'}}\right|\leq \epsilon.$$
Because $n'$ was arbitrary, we know that the inequality above is true for all $n'$. In other words, this proves the statement:
$$\exists M': \forall n'\in \mathbb N: n'>M'\implies |e-b_{n'}|\leq \epsilon$$
Because the choice of $\epsilon$ was arbitrary, we know that the statement above is true for all $\epsilon > 0$, and this is exactly the definition of the limit of the sequence $b_n$ being $e$.

In particular, the changes from your proof are:

*

*There is no splitting of cases when $a_1$ is positive or negative. This splitting of cases is not needed, because in both cases, we will  need $a_n$ to eventually be not only positive, but also larger than some (usually big) number.

*There are no subsequences in my proof, everything follows directly and cleanly from the definitions of limits.

*There is no need for the concept of "shifted" (or "moved", as you call them) sequences.

