Prove that these definitions of $e$ are equivalent Im sorry if this is stupid or obvious, but why 
$$ e=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n$$
AND
$e$ is the unique positive number for which $$ \lim_{h\rightarrow 0}\frac{e^h-1}{h}=1$$?
I mean, how do we know that these two definitions are equivalent? 
Again, maybe it's easy, but I'm just beginning calculus and our teacher just dropped those two definitions...
 A: Expanding the expression $(1+\frac{1}{n})^n$ in the first definition by the binomial theorem gives $\sum_{r=0}^n \binom{n}{r} n^{-r} $, and each term $\binom{n}{r} n^{-r} = \frac{1}{r!} \frac{n (n-1) \cdots (n-r+1)}{n^r}$ at the $n \to \infty$ limit becomes $\frac{1}{r!}$, as the numerator $n (n-1) \cdots (n-r+1)$ of the rational expression on the right has leading term $n^r$, as does the denominator. The identity $e = \sum_{r=0}^\infty \frac{1}{r!}$ can be proved through the theory of Taylor series, and relies on the fact that $\frac{d}{dx} e^x = e^x$. The second definition basically states (by L'Hopital's rule) that $e^x$ has derivative 1 at 0, which is equivalent to $\frac{d}{dx} e^x = e^x$ if you use the further fact that the derivative of any exponential function is proportional to the function itself.
I know this answer uses concepts that you don't know (Taylor series and L'Hopital), but I'm afraid I don't know any more elementary way to make any similar connection. You shouldn't feel stupid: the connections between the two definitions are not obvious.
