TITLE: Combinatorial proof
QUESTION [11 upvotes]: 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
$$e^m=\sum_{k=0}^{\infty}\frac{m^k}{k!}$$

REPLY [2 votes]: $\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:
$$
\expo{-x}\sum_{k = 0}^{\infty}{x^{k} \over k!}=1
$$