In my problem book, there was a question: By defining $e= \lim\limits_{n \to \infty}\left( 1+\frac{1}{n} \right) ^n$ prove that $e^x = \sum\limits_{n=0}^ \infty \frac{x^n}{n!}$. this is a strange question because this book is all about sequence of real numbers and derivatives or Taylor's theorem aren't introduced yet
Here is my attempt:
Lemma 1:- The sequence $e_n := \left( 1+\frac{1}{n} \right) ^n$ converges.
Proof: By binomial theorem $$e_n =1+1 +\frac{1-\frac{1}{n}}{2!} +\frac{(1-\frac{1}{n})(1-\frac{2}{n})}{3!} +\frac{(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})}{4!}+ \dots $$ ,clearly $e_{n+1}> e_n$ then the sequence is strictly increasing. $e_n \leq 2+\sum_{k=2}^ \infty \frac{1}{k!}$ it is obvious that $n!>2^{n-1}$ then $e_n<2+\sum_{k=1}^ \infty \frac{1}{2^k} =3$ by monotone convergence theorem $e_n $ converge to $e:= \sup\{{e_n} \}$
Lemma 2:- the sequence $a_n:= \left( 1-\frac{1}{n} \right)^n $ converges.
Proof: note that $a_n<1$ so the sequence is bounded above, $$\frac{a_{n+1}}{a_n}=\left( \frac{n}{n-1} \right)^\frac{1}{n} \left( \frac{n}{n+1} \right)^{\frac{1}{n+1}} = \left(\left( \frac{n}{n-1} \right)^{n+1} \left( \frac{n}{n+1} \right)^{n} \right)^{\frac{1}{(n)(n+1)}}$$ $$ =\left( \frac{1}{(1-\frac{1}{n^2})(1-\frac{1}{n})}\right)^{\frac{1}{(n)(n+1)}} \geq 1$$ then the sequence is strictly increasing by monotone convergence theorem $a_n $ converge to $a:= \sup\{{a_n} \}$.
Corollary:- the sequence $a_n:=\left( 1-\frac{1}{n} \right)^n$ converges to $\frac{1}{e}$.
Proof: $$a_n=\left(\frac{n-1}{n} \right)^{{n}}$$ Let $k=n+1$ $$a_n=\left(\frac{k}{k+1} \right)^{{k+1}}=\left(\frac{k+1}{k} \right)^{-k-1} $$ $$\lim_{n\to \infty} a_n = \frac{1}{e}$$
Lemma 3 For any monotone increasing sequence $x_n\to \infty$ ,then $\lim\limits_{n\to \infty} \left(1+ \frac{1}{x_n} \right)^{{x_n}}= e$
Proof:- Define $\alpha_n= \lfloor x_n\rfloor $ , $\beta_n=\lceil x_n\rceil $
$$\left(1+ \frac{1}{\alpha_n} \right)^{\alpha_n -1}\leq\left(1+ \frac{1}{x_n} \right)^{{x_n}}\leq \left(1+ \frac{1}{\beta_n} \right)^{\beta_n +1}$$
since $\left(1+ \frac{1}{\alpha_n} \right)^{\alpha_n-1} ,\left(1+ \frac{1}{\beta_n} \right)^{\beta_n +1} $ is a sub sequence of $\left(1+ \frac{1}{n} \right)^{n-1} , \left(1+ \frac{1}{n} \right)^{n+1}$ respectively
then by squeeze theorem $\lim\limits_{n\to \infty} \left(1+ \frac{1}{x_n} \right)^{{x_n}}= e$
Corollary:- $e^x = \lim\limits_{n\to \infty} \left(1+ \frac{x}{n} \right)^{n}$
Proof By lemma 3 $$e^x = \left(\lim\limits_{n\to \infty} \left(1+ \frac{x}{n} \right)^{\frac{n}{x}}\right)^{x}=\lim\limits_{n\to \infty} \left(\left(1+ \frac{x}{n} \right)^{\frac{n}{x}}\right)^{x}=\lim\limits_{n\to \infty} \left(1+ \frac{x}{n} \right)^{n}$$
Here I couldn't rigorously prove that $ \lim\limits_{n\to \infty} \left(1+ \frac{x}{n} \right)^{n}=\sum\limits_{n=0}^ \infty \frac{x^n}{n!}$
Two hints to help you: One, you may have trouble proving $$\lim_{n \rightarrow \infty} (1+ \frac{x}{n})^{\frac{1}{n}} = \lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{x^k}{k!}$$As the left hand side is not $e^x$. In the last corrolary you made some unseen mistake and changed $e^x$s definition (if we want $\frac{x}{n} =\frac{1}{x_n} $, we need $x_n = \frac{n}{x}$).
Second, When you have come to the correct definition of $e^x$, we may find: $$ (1+ \frac{x}{n})^n = 1 + \frac{ {n \choose 1} x}{n} + \ldots + \frac{{ n \choose k} x^k}{n^k}$$ Via the binomial expansion. And for large $n$, $n^k$ actually dissapears when combined the $n \choose k$, which you can read about here