Can someone please explain $e$ in layman's term? I never really understood what $e$ means and I'm always terrified when I see it in equations. What is it? Can somebody dumb it down for me? I know it's a constant. Is it as simple as that?
 A: Euler's Number, $e\approx2.718281828$, is a special constant in calculus because of several important properties it holds. 
First, $y=e^{x}$ (also written as $y=\exp(x)$ due to the exponential growth of the function) and constant multiples thereof is the only function for which:
$$\frac{dy}{dx}=y$$
Which means in simple language that the gradient of the curve at a point is proportional to the value of the curve at that point. For instance, at the point $(1,e)$, the gradient of the tangent (shown as blue in the graph below) to the curve $y=e^{x}$ (shown in red) is $e$.
             
             
        

A proof that $y=C\exp(x)$ is the only real function for which this rule applies can be found here.
Euler's Number is also the number which the expression $\left(1+\frac{1}{x}\right)^{x}$ approaches as $x$ gets closer and closer to infinity. Formally:
$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e$$
Euler's number also can be found as the unique positive real number $e$, such that:
$$\int_{1}^{e}\frac{dt}{t}=1$$
Which in common language means the area under the curve $y=\frac{1}{x}$ between $1$ and $e$ has unit area (or area equal to 1).
Also, interestingly, $e$ can be found by applying Stirling's approximation to the factorial in reverse to get:
$$e=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}$$
A: $e$, the constant, is the limit of $(1 + {1\over n})^n$ as n approaches infinity; As expressed by the series: $$e = 1 + {1\over 1} + {1\over 1\cdot2 } + {1\over 1\cdot2\cdot3} + {1\over 1\cdot2\cdot3\cdot4} + ... + {1\over 1\cdot 2\cdot...\cdot\infty}$$
It is the base of the natural logarithm.
A: The idea of $e$ may be explained through the concept of compound interest.  Let's say that you earn an annual interest rate $r$ on a principal $P$ dollars.  Let's say that the money is compounded annually.  Then after $n$ compounding periods, you  have $P (1+r)^n$ dollars.
Now lets say we compound semiannually; then after the same amount of time $n$ years, i.e., $2 n$ compounding periods, you will have 
$$P \left ( 1+\frac{r}{2} \right )^{2 n}$$
dollars.  You can now imagine more frequent compounding events; let's say that there are $M$ such events per year.  Then after $n$ years you will have
$$P \left ( 1+\frac{r}{M} \right )^{M n}$$
dollars.
Now imagine that you are compounding continuously (like, every microsecond).  This corresponds to the limit as $M \to \infty$: after $n$ years, you will have
$$P \lim_{M \to \infty} \left ( 1+\frac{r}{M} \right )^{M n} =P\, e^{r n}$$
dollars.
A: The simplest way to understand it is, consider the following equation:
$$f(x)=\left(1+\frac{1}{x}\right)^x$$
As $x$ gets larger and larger, notice what number it gets closer to:
$$f(1)=2$$
$$f(2)=2.25$$
$$f(3)\approx 2.37073$$
$$...$$
$$f(1000)\approx 2.7169$$
$$...$$
It is important in mathematics because it is arguably the single real number most commonly found in nature (from heat flow to battery discharge to population growth/decay models), which is why its inverse is called the natural logarithm.
A: On can show that there are functions $f(x)$ with the property 
$$\tag1f'(x)=f(x)$$ for all $x$. If one additionally requires $f(0)=1$, the solution is unique and called $\exp(x)$. From this property alone, it follows that the general solution to $f'(x)=f(x)$ is $f(x)=c\exp(x)$ with $c=f(0)$. And also that $$\tag2\exp(x+y)=\exp(y)\exp(x)$$ for all $x,y$
because $x\mapsto \exp(x+y)$ is also a solution of $(1)$. Especially, $\exp(n)=\exp(1)\exp(1)\ldots\exp(1)=(\exp(1))^n$. In effect, because of $(2)$, the function $x\mapsto \exp(x)$ suspiciously behaves like exponentiation and is therefore usually suggestively written like exponentiation, i.e. we write $e^x$ instead of $\exp(x)$. Of course for this to make sense, we must define $e=\exp(1)$. It turns out that $\exp(1)\approx 2.71828$.
