Combinatorial proof Using notion of derivative of functions from Taylor formula follow that
$$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$
Is there any elementary combinatorial proof of this formula
here is my proof for $x$ natural number
Denote by $P_k^m$ number of $k$-permutations with unlimited repetitions of elements from a $m$-set then we can prove that
$$P_k^m=\sum_{r_0+r_1+...+r_{m-1}=k}\frac{k!}{r_0 !...r_{m-1}!}$$
also is valid
$$P_k^m=m^k$$
Based on first formula we can derive that
$$\sum_{k=0}^{\infty}P_k^m\frac{x^k}{k!}=\left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)^m$$
from second formula
$$\sum_{k=0}^{\infty}P_k^m\frac{x^k}{k!}=\sum_{k=0}^{\infty}\frac{(mx)^k}{k!}$$
now is clear that 
$$\sum_{k=0}^{\infty}\frac{(mx)^k}{k!}=\left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)^m$$
from last equation for $x=1$ taking in account that
$$\sum_{k=0}^{\infty}\frac{1}{k!}=e=2,71828...$$
we have finally that for natural number $m$ is valid formula
$$e^m=\sum_{k=0}^{\infty}\frac{m^k}{k!}$$

We will handle $x>0$ here.
If we define $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$, then $e^x=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}$.  Note that since $0\le nx-\lfloor nx\rfloor<1$,
$$
\begin{align}
e^x&=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}\\
   &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor} \left(1+\frac{1}{n}\right)^{nx-\lfloor nx\rfloor}\\
   &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor} \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx-\lfloor nx\rfloor}\\
   &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor}
\end{align}
$$
Using the binomial theorem,
$$
\begin{align}
\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor}
&=\sum_{k=0}^{\lfloor nx\rfloor} \frac{1}{k!}\frac{P({\lfloor nx\rfloor},k)}{n^k}\\
&=\sum_{k=0}^\infty \frac{1}{k!}\frac{P({\lfloor nx\rfloor},k)}{n^k}
\end{align}
$$
Where $P(n,k)=n(n-1)(n-2)...(n-k+1)$ is the number of permutations of $n$ things taken $k$ at a time.
Note that $0\le\frac{P({\lfloor nx\rfloor},k)}{n^k}\le x^k$ and that $\sum_{k=0}^\infty \frac{x^k}{k!}$ converges absolutely.  Thus, if we choose an $\epsilon>0$, we can find an $N$ large enough so that, for all $n$,
$$
0\le\sum_{k=N}^\infty \frac{1}{k!}\left(x^k-\frac{P({\lfloor nx\rfloor},k)}{n^k}\right)\le\frac{\epsilon}{2}
$$
Furthermore, note that $\lim_{n\to\infty}\frac{P({\lfloor nx\rfloor},k)}{n^k}=x^k$. Therefore, we can choose an $n$ large enough so that
$$
0\le\sum_{k=0}^{N-1} \frac{1}{k!}\left(x^k-\frac{P({\lfloor nx\rfloor},k)}{n^k}\right)\le\frac{\epsilon}{2}
$$
Thus, for n large enough,
$$
0\le\sum_{k=0}^\infty \frac{1}{k!}\left(x^k-\frac{P({\lfloor nx\rfloor},k)}{n^k}\right)\le\epsilon
$$
Therefore,
$$
\lim_{n\to\infty}\;\sum_{k=0}^\infty\frac{1}{k!}\frac{P({\lfloor nx\rfloor},k)}{n^k}=\sum_{k=0}^\infty\frac{x^k}{k!}
$$
Summarizing, we have
$$
\begin{align}
e^x&=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}\\
   &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor}\\
   &=\lim_{n\to\infty}\;\sum_{k=0}^\infty \frac{1}{k!}\frac{P({\lfloor nx\rfloor},k)}{n^k}\\
   &=\sum_{k=0}^\infty\frac{x^k}{k!}
\end{align}
$$
