Could you explain why $\frac{d}{dx} e^x = e^x$ "intuitively"? As the title implies, It is seems that $e^x$ is the only function whoes derivative is the same as itself.
thanks.

One interesting way of looking at this is by turning the differential equation into an integral equation. 
The problem $ y'=y $ is the same as $ \int y \ dx = y $. For the problem to be well defined I have to define an initial condition, we use the condition $y(0)=1$.
Now I am going to guess an answer for y. The simplest function $y(x)$ that I know which goes through $(0,1)$ is $y_0(x)=1$. The subscript here has no special mathematical meaning, it just signifies that this is my first guess.
If we put our guess into the integral equation we get, $$ \int y_0(x) \ dx = \int 1 \ dx = x + C . $$
The integral added a constant that wasn't there before, our guess must have not been very good so we will take this new result as our second guess hoping that it knows more about the problem. So our new guess is,
$$y_1(x) = x + C = x + 1 $$
Notice that we must set C=1 so that $y_1(0)=1$. We will substitute this into the integral equation again which will give us a new guess $y_2(x)$.
$$ y_2(x) = \int y_1(x) \ dx = \int x + 1 \ dx = \frac{x^2}{2} + x + C = \frac{x^2}{2} + x + 1 $$
If we do this a few more times we will get the following,
$$ y_3(x) = \frac{x^3}{6} + \frac{x^2}{2} + x + 1 $$
$$ y_4(x) = \frac{x^4}{24} + \frac{x^3}{6} + \frac{x^2}{2} + x + 1 $$
If you look closely at the coefficients you will see that we are generating the Maclaurin series for $e^x$. 
This is happening because the only function which is it's own anti-derivative is the exponential function. In the terminology of dynamics we would say that the infinite series representation of $e^x$ is a fixed point of the integral operator. It is also an attractor which means that repeated integration of $some$ unrelated functions will produce a sequence of functions that approach $e^x$.
