How can I prove $\lim \limits_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e$ without involving function limit?

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.

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 $$\left(1+\frac{k+1}{n}\right)^n=\left(\frac{n+k+1}{n}\right)^n=\left(1+\frac{k}{n}\right)^n\left(1+\frac{1}{n+k}\right)^n=\underbrace{\left(1+\frac{k}{n}\right)^n}_{\to e^k\ (hyp\ induction)}\underbrace{\left(1+\frac{1}{n+k}\right)^{n+k}}_{\to e}\underbrace{\left(1+\frac{1}{n+k}\right)^{-k}}_{\to 1}\underset{n\to\infty }{\longrightarrow }e^{k+1}$$