Is there a direct way to prove $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n=\sum\limits_{n=0}\frac{1}{n!}$? How to prove the equality of the two definitions of $e$ directly?
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\sum_{n=0}\frac{1}{n!}$$
In other words, I want to know a proof that is not &quot;prove $e^x=\sum\limits_{n=0}^\infty\frac{1}{n!}x^n$ and then substitute $x=1$&quot;. I guess we do not need $e^x$ to prove this equality.

There are answers in other questions like this:
$$\begin{aligned}
\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n &amp;= \lim_{n\to\infty}\sum_{k=0}^n \frac{1}{k!} \prod_{m=0}^{k-1}\left(1-\frac{m}{n}\right)\\
&amp;=^?\sum_{k=0}^\infty\left[\frac{1}{k!}\lim_{n\to\infty}\prod_{m=0}^{k-1}\left(1-\frac{m}{n}\right)\right]\\
&amp;=\sum_{k=0}^\infty\frac{1}{k!}
\end{aligned}$$
However, I think this operation of limits is wrong. For example, $\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\frac{1}{n}=1$ and $\sum\limits_{k=1}^\infty\lim\limits_{n\to\infty}\frac{1}{n}=0$.

None of the answers to Prove the definitions of e to be equivalent answers this question. There are 3 answers to it:

https://math.stackexchange.com/a/69825/769808
https://math.stackexchange.com/a/69809/769808
https://math.stackexchange.com/a/69957/769808

1 is begging the question. It assumes $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$.
2 is wrong. The operation of limits is incorrect.
3 is not direct proof.
A nice direct proof comes from Rudin:
Let
$$
s_n=\sum_{k=0}^n\frac{1}{k!}\;,\;\;t_n=\left(1+\frac{1}{n}\right)^{n}
$$
First, $s_n$ converges, as it is an increasing sequence that is bounded above by $1 + \sum_{k=0}^{n-1}\frac{1}{2^k} \le 3$.
Let $e:=\lim_{n\rightarrow\infty}s_n$. By the binomial theorem,
$$
\small{t_n = 1 + 1+\frac{1}{2!}\left(1-\frac{1}{n}\right) + \frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) +\; \cdots \;+  \frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)}
$$
Hence, $t_n\le s_n$, so that $$\limsup_{n\rightarrow\infty}t_n\le e$$
Next, if $n\ge m$,
$$
t_n\ge 1 + 1 + \frac{1}{2!}\left(1-\frac{1}{n}\right)+ \cdots +  \frac{1}{m!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)
$$
Let $n\rightarrow\infty$, keeping $m$ fixed. We get
$$
\liminf_{n\rightarrow\infty}t_n\ge 1+1+\frac{1}{2!} + \cdots + \frac{1}{m!}
$$
so that $s_m\le\liminf_{n\rightarrow\infty} t_n$. Letting $m\rightarrow\infty$, we finally get:
$$
e\le\liminf_{n\rightarrow\infty}t_n
$$
In the original series you can enter with $\lim$ under summation because it is dominated by $\sum_{k=1}^\infty\frac{1}{k!}$ which is convergent. Read more about domination here: https://en.wikipedia.org/wiki/Dominated_convergence_theorem (and the simplification for series: Discrete version of dominated convergence thm ).
But yes, this is not as trivial as other answers present. I'm not aware of an easier solution.