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
$$e^m=\sum_{k=0}^{\infty}\frac{m^k}{k!}$$
 A: Fact: Every nonzero continuous function $F\colon\mathbb R\to\mathbb R_+$ where $F(a+b)=F(a)F(b)$ is an exponential function for some base: $F(x)=a^x$. To prove this, first note that $F(0)=F(0+0)=F(0)F(0)$. So $F(0)^2=F(0)$. So $F(0)\in\{0,1\}$. But if $F(0)=0$, the whole function is identically zero.
Now let $a=F(1)$ and first prove this for positive integers $x=n$: $F(n)=F(1+\cdots+1)=F(1)\cdots F(1)=a^n$. Extend to negative integers using $1=F(0)=F(n-n)=F(n)F(-n)=a^nF(-n)$. So $F(-n)=a^{-n}$. Now extend to rationals $x=p/q$: $a^p=F(p)=F(q\cdot p/q)=F(p/q)^q$. So $F(p/q)=a^{p/q}$.  To extend to all reals, we invoke continuity. Given a real $r$, find a sequence of rationals that converge to it $q_n\to r$. Then $F(r)=\lim_{n\to\infty}F(q_n)=\lim_{n\to\infty}a^{q_n}=a^r$. This last step was not combinatorial, so I am cheating.
Now you can combinatorially prove that if you let $F(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$, then $F(a+b)=F(a)F(b)$. So $F(x)=a^x$ for some $a>0$. Now you just need to show $a=e$. This will depend on how you defined $e$ in the first place. I'll assume $e=\lim_{n\to\infty}(1+1/n)^n$. Apply the binomial theorem to $(1+1/n)^n$ to get
$$(1+1/n)^n=\sum_{k=0}^n \binom{n}{k}n^{-k}=1+n/n+\frac{n(n-1)}{2n^2}+\cdots$$
This is approximately $1+1+\frac{1}{2}+\frac{1}{3!}+\cdots.$ Taking the limit as $n$ goes to infinity, we get $e=\sum_{k=0}^\infty \frac{x^k}{k!}$. (This step can be made rigorous. I've given a sketch.) But $F(1)=a^1$ matches this expression, so $a=e$.
A: I don't know if this is what you are looking for, but working from the fact that $\frac{d}{dx}e^x=e^x$, we can say that if
$$
e^x=\sum_{k=0}^\infty\;a_k\;x^k
$$
then, by taking the derivative of both sides, we get
$$
\begin{align}
e^x&=\sum_{k=1}^\infty\;k\;a_k\;x^{k-1}\\
   &=\sum_{k=0}^\infty\;(k+1)\;a_{k+1}\;x^k
\end{align}
$$
Equating the coefficients of $x^k$ in these two formulas for $e^x$, we get that $a_k=\frac{1}{k}\;a_{k-1}$.  Since $e^0=1$, we have that $a_0=1$.  Thus, we are lead to the conclusion that $a_k=\frac{1}{k!}$.  That is,
$$
e^x=\sum_{k=0}^\infty\frac{x^k}{k!}
$$
A: $\newcommand{\+}{^{\dagger}}
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Assume you flip $N$ times a coin. Assume that he probability to have tail is
$x/N$ and head is $1 - x/N$. The probability of $k$ tails is
$$
{N \choose k}\pars{x \over N}^{k}\pars{1 - {x \over N}}^{N - k}
$$
The probability of "no tails" or $1$ tail or $2$ tails $\ldots$ or $N$ tails is obviously equal to $1$. Namely:
$$
\sum_{k = 0}^{N}{N \choose k}\pars{x \over N}^{k}\pars{1 - {x \over N}}^{N - k}
=1
$$
Now, take the limit $N \to \infty$ ( a delicate and exquisit one !!! ) and the result is:
$$
\expo{-x}\sum_{k = 0}^{\infty}{x^{k} \over k!}=1
$$
A: 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
$$
\begin{align}
e^x&=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}\\
   &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor}\\
   &=\lim_{n\to\infty}\;\sum_{k=0}^\infty \frac{1}{k!}\frac{P({\lfloor nx\rfloor},k)}{n^k}\\
   &=\sum_{k=0}^\infty\frac{x^k}{k!}
\end{align}
$$
