<s>[INST] How to prove that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$? How to prove that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ using the fact that $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$ ?
So,
$$e^x=\lim_{n\to\infty }\sum_{k=0}^n \binom{n}{k}\left(\frac{x}{n}\right)^k=\lim_{n\to\infty }\sum_{k=0}^n\frac{n!}{(n-k)!n^k}\frac{x^k}{k!}.$$
I think that I have to prove that $\frac{n!}{(n-k)!n^k}=1$ but I didn't have success.
 [/INST] 1. Solution using $\,e^{\,x} = \dfrac{d}{dx}\,e^{\,x}\,$ property of exponent
Assume that the exponent function can be represented as a series with unknown coefficients:
$$
e^{\,x} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots = \sum_{n=0}^{\infty} a_n x^n
$$
Recall the fundamental property of exponent $\, \dfrac{d}{dx} \big(e^{\,x} \big)  = e^{\,x}$.
Applying this property to the series for of exponent, we get
\begin{align}
\dfrac{d}{dx} \,e^{\,x}  = e^{\,x}
& \implies 
\dfrac{d}{dx} \left(\sum_{n=0}^{\infty} a_n x^n \right)  = \sum_{n=0}^{\infty} a_n x^n
\\ & \implies 
\dfrac{d}{dx} \big( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\big) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots
\\ & \implies 
0  +   a_1 + 2 \, a_2\, x + 3 \, a_3\, x^2 + \ldots  =  a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots
\\ & \implies 
\sum_{n=1}^{\infty}n \,a_n\,x^{n-1}  =  \sum_{n=0}^{\infty} a_n x^n
\iff
\sum_{n=0}^{\infty}\left(n+1\right) a_{n+1}\,x^{n}  =  \sum_{n=0}^{\infty} a_n x^n
\\ & \implies 
\left(n+1\right)a_{n+1}  = a_n
\\ &
\implies
a_{n+1}  = \frac{a_n}{n+1}
\end{align}
The last equation  can be rewritten as $\,a_{n}  = \dfrac{1}{n}a_{n-1}, \,$ so that
$$
a_{n}  = \frac{1}{n}\,a_{n-1} =  \frac{1}{n}\,\frac{1}{n-1} \,a_{n-2}= 
\frac{1}{n}\,\frac{1}{n-1}\,\frac{1}{n-2}\,a_{n-3} = \ldots = 
\frac{1}{ n!}\,a_0\tag{1.1}
$$
Observer that $\,e^0 = 1,\,$  so we can write
$$
e^{\,x}\Big\rvert_{x=0} = 
\big( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\big)\Big\rvert_{x=0} 
= a_0 = 1
$$
This fact combined with equation $(1.1)$ gives us the explicit expression for coefficient $\,a_n =  \dfrac{1}{n!}.\,$
Therefore we finally write 
$$
\bbox[4pt, border:2.5pt solid #FF0000]{\ \ e^{\,x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}\ \,}
$$
Q.E.D.

EDIT: As requested in comments, here I provide solution using $\,\displaystyle e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ expression.

2. Solution using $\,\displaystyle e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ property of exponent
I do not believe that it is possible (at least within reasonable timespan) express the exponentsas  $\,e^{\,x}=\sum_{k=0}^\infty \frac{x^k}{k!}\,$ using only  algebraic operations and the expression formula $\, e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ as starting point.
However, it is possible to show the equivalence of these two definitions of $\,e^{\,x}\,$  as $\,n\to \infty.\,$
Indeed, for any $x\ge 0$ let us define 
$$
S_n = \sum_{k=0}^n \frac{x^k}{k!}, \qquad 
L_n = \left(1+\frac{x}{n}\right)^n.
$$
Then, by Newton's binomial
$$
\begin{aligned}
L_n & = \sum_{k=0}^n {n\choose k} \,\frac{x^k}{n^k}
= 1 + x + \sum_{k=2}^{n} 
\frac{n\cdot \left(n-1\right)\cdot  \left(n-2\right)\cdot \ldots\cdot  \left(n-(k-1)\right)}{k! \,n^k}=
\\ & =
1 + x 
+ \frac{x^2}{2!}\,\left(1 - \frac{1}{n} \right) 
+ \frac{x^3}{3!}\,\left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) 
+ \ldots
+ \frac{x^n}{n!}\,\left(1 - \frac{1}{n} \right) \cdots 
\left(1 - \frac{n-1}{n} \right) \leq S_n
\end{aligned}
$$
Therefore 
$$
\limsup_{n\to\infty}L_n \leq \limsup_{n\to\infty}S_n = e^{\,x}.\tag{2.1}
$$
On the other hand, for any positive integer $\, m\,$ such that $\,2\le m \le n\,$  we have
$$
1 + x 
+ \frac{x^2}{2!}\,\left(1 - \frac{1}{n} \right) 
+ \ldots
+ \frac{x^m}{m!}\,\left(1 - \frac{1}{n} \right)\left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right) 
\le L_n
$$
If we fix $\,m\,$ and let $\,n\to\infty,\,$ then we get
$$
S_m = 1 + x + \frac{x^2}{2!} + \ldots + \frac{x^m}{m!}
\leq \liminf_{n\to\infty}L_n\tag{2.2}
$$
Letting $\,m\to\infty\,$ in inequality $(2.2)$ and combining it with inequality $(2.1)$, we get
$$
e^{\,x} = \limsup_{n\to\infty}L_n\leq \lim_{n\to\infty} S_n \leq \liminf_{n\to\infty}L_n = e^{\,x}
$$
and thus 
$$
\bbox[5pt, border:2.5pt solid #FF0000]{\lim_{n\to \infty}S_n = \sum_{n=0}^\infty \frac{x^n}{n!}=e^{\,x}}
$$
</s>