Bernoulli's representation of Euler's number, i.e $e=\lim \limits_{x\to \infty} \left(1+\frac{1}{x}\right)^x $ 
Possible Duplicates:
Finding the limit of $n/\sqrt[n]{n!}$
How come such different methods result in the same number, $e$? 

I've seen this formula several thousand times: $$e=\lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x $$
I know that it was discovered by Bernoulli when he was working with compound interest problems, but I haven't seen the proof anywhere. Does anyone know how to rigorously demonstrate this relationship?
EDIT:
Sorry for my lack of knowledge in this, I'll try to state the question more clearly. How do we prove the following?
$$ \lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x = \sum_{k=0}^{\infty}\frac{1}{k!}$$
 A: It does rather matter how you want to define $e$ in the first place. One way to define $e$ is to prove that the sequence whose $n$-th term is $(1 + \frac{1}{n})^n$ is increasing, but bounded above, and therefore converges to its least upper bound,
which may be defined as $e$. More generally, we may define $e^x$ as
$\lim_{n \to \infty} (1 + \frac{x}{n})^n$ for any real $x$ ( and the limit always exists). Then it's easy to verify from this definition that $e^{x+y} = e^{x}.e^{y}$
for all $x,y \in \mathbb{R}$. With this approach the Bernoulli representation of $e$
is almost a non-issue. 
The very definition  $(1 + \frac{1}{x})^{x}$ for non-integral $x$ (as $\exp(x \log(1 + \frac{1}{x}))$),  presupposes that $e$ (and the natural logarithm) have already been defined.
Another way to define the function $e^x$ from first principles, adopted, for example, in Spivak's "Calculus"), is as the inverse function of the logarithm, where
$\log(x)$ is defined as $\int_{1}^{x}\frac{1}{t} dt$ for $x >0$. Then the fundamental theorem of Calculus gives $\log'(x) = \frac{1}{x}$ for $x >0$, and if we define
the exponential function as the inverse of the logarithm function, it is its own derivative. Since this function is always positive, the exponential function is increasing everywhere. The mean value theorem tells us that $x\log(1 + \frac{1}{x})
= \frac{1}{\theta}$ for some $\theta \in (1,1+\frac{1}{x})$ when $x >0.$
As $x \to \infty$, we see that $\theta \to 1$. Since $e^{x}$ is differentiable
everywhere, it is certainly continuous, so that as $x \to \infty$, 
$\exp(x \log(1 + \frac{1}{x})) \to \exp(1) = e.$
NOTE ADDED: Since the question has been rephrased taking $e = \sum_{i=0}^{\infty} \frac{1}{i!}$ after the above was written, I add that the second approach here does that, since the fact that the exponential function is its own derivative shows that its 
Maclaurin series is the expected $e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, and that 
this converges for all real $x$ using the standard form for the remainder in Taylor's theorem (as, eg, in Spivak's book). 
A: Take a look at http://www.whim.org/nebula/math/eseries.html.  On that page, it is shown that $\left(1+\frac{1}{n}\right)^n\le e$. Then, it is shown, using the binomial theorem, that $e-\left(1+\frac{1}{n}\right)^n$ can be made as small as desired by choosing $n$ large enough.
Update:
If the link above does not work, the content of that link has been moved to this answer.
A: Well, the problem with $e$ is that there are many different ways of defining it. But this is another way.
Suppose the limit exists, and call it $L$.
$\log L = \lim x \log \left( \dfrac{x + 1}{x} \right) = \lim \dfrac{ \log \frac{x + 1}{x}}{\frac{1}{x}} = \lim \dfrac{ \frac{x}{x+1} \cdot (-x^{-2} \cdot (x + 1) + x^{-1})}{ -x^{-2} } = \; \;...$
$... \;= \lim \dfrac{ \frac{x}{x+1} \cdot \frac{-1}{x^2} }{\frac{-1}{x^2}} = 1$
So $L = e^1$.
For a different approach.
A: From the binomial theorem
$$\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^n {n \choose k} \frac{1}{n^k} = \sum_{k=0}^n \frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\cdots\frac{n-k+1}{n}\frac{1}{k!}$$
but as $n \to \infty$, each term in the sum increases towards a limit of $\frac{1}{k!}$, and the number of terms to be summed increases so 
$$\left(1+\frac{1}{n}\right)^n \to  \sum_{k=0}^\infty \frac{1}{k!}.$$
A: We know that 
$\ln\left(\left(1+\frac{1}{n}\right)^n\right)= n \ln \left(1 + \frac{1}{n}\right)$
Now suppose that $x = \frac{1}{n}$. Thus,
$n \ln \left(1 + \frac{1}{n}\right) = \displaystyle \left(\frac{1}{x} \ln\left( 1 + x\right)\right) = \displaystyle \left(\frac{1}{x} (\ln\left( 1 + x\right) - \ln 1)\right)$
Now if we send $x$ to 0, we see that $\displaystyle\lim_{x \rightarrow 0}\left(\frac{\ln\left( 1 + x\right) - \ln 1}{x}\right)$, which equals the derivative of $\ln$ at x = 1, which is 1.
Thus, we know that $\displaystyle\lim_{n \rightarrow \infty}\left(\ln\left(\left(1+\frac{1}{n}\right)^n\right)\right) = \lim_{x \rightarrow 0}\left(\frac{\ln\left( 1 + x\right) - \ln 1}{x}\right) = 1$
Now, using the fact that $e^{\ln x} = x$, we know that $\lim_{n \rightarrow \infty}\left(\left(1 + \frac{1}{n}\right)^n\right) = e^{\ln\left(\lim_{n \rightarrow \infty}\left(\left(1 + \frac{1}{n}\right)^n\right)\right)} = e^{\lim_{n \rightarrow \infty}\left(\ln\left(\left(1 + \frac{1}{n}\right)^n\right)\right)} = e^{1} = e$. 
A: $$\mathrm {Log}\left(\displaystyle\lim_{x\rightarrow\infty} \left(1 + \frac{1}{x}\right)^{x}\right) = \displaystyle\lim_{x\rightarrow 0}\text{ } \mathrm {Log}  \left((1 + x)^{\frac{1}{x}}\right) =  \lim_{x\rightarrow 0} \frac{\mathrm {Log}(1+x)}x = \lim_{x\rightarrow 0} \text{ }\displaystyle\sum_{i=0}^{\infty} \frac{x^i}{i+1} = 1.$$
