How was Euler's number defined using logarithmic differentiation? Consider a well known definition of e: the limit of $(1+ 1/n)^n$ as $n$ approaches infinity.
I was under the impression that this was the first formula giving the exact value of e. This would make sense as Euler first observed the number (as far as I know) in compound interest computations, and this formula is in fact computing continuously compounded interest.
If I am in error, please help guide me toward truth.
If not, then here is my question:
Assuming I have not overlooked a simpler method of computing the limit, we must take the natural log of the limit before employing L'Hopital's rule and exponentiating to ultimately yield e... 
But e is in the definition of the natural log, so wouldn't this be a circular argument to show that this limit is equal to e?
 A: Obviously, if you use this limit to define $e$, then, by definition, it is $e$, as long as you've shown the limit exists. You can't prove it is $e$ if you don't have $e$ defined.
We can show the limit exists.
A typical elementary way is to show that $a_n=(1+1/n)^n$ is increasing and has upper bound $3.$ Thus a limit exists.
Increasing means $a_{n+1}> a_n$ or $$\frac{(n+2)^{n+1}}{(n+1)^{n+1}}>\frac{(n+1)^n}{n^n}$$ 
or $$\left(1-\frac{1}{(n+1)^2}\right)^{n+1}>\frac{n}{n+1}$$
which follows from Bernoulli's inequality.
The upper bound can be seen by noting that:
$$\left(1+\frac1n\right)^n=\sum_{k=0}^{n}\frac{1}{n^k}\binom{n}{k}<\sum_{k=0}^{n}\frac{1}{k!}<1+1+\sum_{i=1}^{\infty}\frac{1}{2^i}=3$$
That the value $e$ has such lovely properties is another matter, but we can actually start with defining $$\exp(x)=\lim_{n\to\infty}\left(1+\frac xn\right)^n,$$ first showing the limit exists and then proving via elementary means that $\exp(x)\exp(y)=\exp(x+y)$ and other properties. 
A: There are two common definitions of $e$
$e = \lim_\limits{n\to\infty} (1+\frac {1}{n})^n$
and 
$e = \lim_\limits{n\to\infty} \sum_\limits{k=0}^n \frac {1}{k!}$  
The first one comes from the compound interest calculation, and is actually the older to the two.
The second is usually first encountered with the Taylor series of $e^x$ but is frequently more useful.  Certainly it is easier to use to find an aproximation for $e.$
Are these two definitions equal?
Do a binomial expansion on $(1+\frac {1}{n})^n$
$1 + 1 + \frac {n-1}{2n} + \frac {(n-1)(n-2)}{6n^2} + \cdots +\frac{(n-1)!}{n!n^{n-1}}$
$1 + 1 + \frac {1}{2!}(1-\frac {1}{n}) + \frac {1}{3!}(1-\frac {1}{n})(1-\frac {2}{n}) + \cdots + \frac {1}{n!} (1 - \frac {1}{n})\cdots(1-\frac {n-1}{n}) \le \sum_\limits{k=0}^n \frac {1}{k!}$ 
Choose $m<n$
$1 + 1 + \frac {1}{2!}(1-\frac {1}{n}) + \cdots + \frac {1}{m!} (1 - \frac {1}{n})\cdots(1-\frac {m-1}{n})\le (1+\frac {1}{n})^n \le \sum_\limits{k=0}^n \frac {1}{k!}$ 
Keeping $m$ fixed let $n$ go to infinity.
$1 + 1 + \frac {1}{2!} + \frac {1}{3!} + \cdots + \frac {1}{m!}\le  \lim_\limits{n\to\infty} (1+\frac {1}{n})^n\le \lim_\limits{n\to\infty} \sum_\limits{k=0}^n \frac {1}{k!}$ 
and now let $m$ approach infinity.
$\lim_\limits{n\to\infty} (1+\frac {1}{n})^n = \lim_\limits{n\to\infty} \sum_\limits{k=0}^n \frac {1}{k!}$
