TITLE: Why is $ \sum_{n=0}^{\infty}\frac{x^n}{n!} = e^x$?
QUESTION [8 upvotes]: I am trying to see where this relationship comes from: $\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!} = e^x$ Does anyone have any special knowledge that me and my summer math teacher doesn't know about? He went over generating functions a few days ago and I did not know exactly what he was talking about. In the middle of his lecture he wrote this generating function. He explained that in this case $a_{n} = \frac{1}{n!}$ for $\displaystyle \sum_{n=0}^{\infty} a_{n} x^n$. 
Is the relationship a definition or can it be derived using $\displaystyle \sum_{n=0}^{\infty} x^n $?

REPLY [4 votes]: Another approach: I imagine you know that
$$\lim_{n \to \infty} (1+x/n)^n = e^x.$$
Expanding by the binomial theorem,
$$(1+x/n)^n = 1+n \frac{x}{n} + \frac{n(n-1)}{2} \frac{x^2}{n^2} + \frac{n(n-1)(n-2)}{3!} \frac{x^3}{n^3} + \cdots$$
$$=1+x+(1-1/n) \frac{x^2}{2}+(1-1/n)(1-2/n) \frac{x^3}{3} + \cdots$$
As $n \to \infty$, the $k$th term goes to $x^k/k!$. One still has to do some work to turn this into a proof, as you can't just assume willy-nilly that the limit of an infinite sum will be the sum of the limits, but it is at least very suggestive, and can be made into a completely correct argument.