How to show $\sum_{x=0}^{\infty }{\frac{a^{x}}{x!}\; =\; e^{a}}$ I've got a basic problem here with deriving the poisson distribution, where in part this summation is needed to show the expectation of the distribution is a parameter of the distribution's function, but while I can see from the derivation that this is the end result, I cannot see how it would be.
$$\sum_{x=0}^{\infty }{\frac{a^{x}}{x!}\; =\; e^{a}}$$
If I expand this, I get:
$$\lim_{n \to \infty} \frac{a^{0}}{0!}+\frac{a^{1}}{1!}+\frac{a^{2}}{2!}+...+\frac{a^{n}}{n!}$$
...but I'm not seeing it beyond this. Any pointers?
 A: If you define the exponential function as $$e^z=\lim_{n\to\infty}\left(1+\frac {z} {n}\right)^n$$
Then $$e^z=\lim_{n\to\infty}\left(1+\frac {z} {n}\right)^n=\lim_{n\to\infty}\sum_{k=0}^n\frac1{n^k}{n\choose k}z^k$$
Now 
$$\lim_{n\to\infty}\frac{1}{n^k}{n\choose k}=\lim_{n\to\infty}\frac{1}{n^k}\frac{n!}{k!(n-k)!}=\frac{1}{k!}\lim_{n\to\infty}\frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k}=\frac1{k!}$$
The limit equals $1$ because both the top and bottom of the fraction are polynomials of degree $k$, so the limit is the ratio of their leading coefficients ($1$ and $1$). Hence
$$e^z=\lim_{n\to\infty}\left(1+\frac {z} {n}\right)^n=\lim_{n\to\infty}\sum_{k=0}^n\frac1{n^k}{n\choose k}z^k=\sum_{k=0}^\infty\frac{z^k}{k!}$$
A: As an addition (maybe not very necessary here) to @Paul's answer:
Provided you definition of $\exp$ is (otherwise you may just skip my answer)
$$e:=\lim_{n\to\infty}\Big(1+\frac1n\Big)^n$$
which is more common as far as I know. Then $\forall z\in \Bbb R^+$, we have
$$\lim_{n\to\infty}\Big(1+\frac zn\Big)^n=\lim_{n/z\to\infty}\Big[(1+\frac{1}{n/z})^{n/z}\Big]^z=\Big[\lim_{n/z\to\infty}(1+\frac{1}{n/z})^{n/z}\Big]^z=:e^z$$
And then for the $z<0$ case, note that
$$\lim_{n\to\infty}\Big(1-\frac1n\Big)^n=\lim_{n\to\infty}\Big(1-\frac{1}{n+1}\Big)^{n+1}=\lim_{n\to\infty}\Big(1+\frac1n\Big)^{-1}\frac{1}{\Big(1+\frac1n\Big)^n}=:\frac1e$$
Take it from here, and you should be able to show $\forall z\in\Bbb R^{-}$
$$\lim_{n\to\infty}(1+\frac zn)^n=:e^z$$
And $z=0$ case is too trivial for discussion here.
The rest is all in @Paul's answer.
