infinite derivative of $e^x$ i have started  thinking about  one topic  a few   days ago and    i am confused   if i am wrong  or what happens,generally we know that function $e^x$  is somehow  'magic',which means  that derivative  and integral of this function is  again $e^x$(let   reject  constant term during the integral).but on the other hand  we can say that (here let's use d  as  sign of derivative)
$$d(e^x)=d(e*e^{x-1})$$
which is equal to  $d(e)*e^{x-1}+ e*d(e^{x-1})$
because of  feature  of derivative of two function(in this case our function $f(x)=e$ is constant),clearly first term is zero,so we have  $e*d(e^{x-1})$,if we continue it  to  infinite time,  we can see that in derivative  sign    power approaches  $x$,or something like this
$$d(e^{x-1}),d(e^{x-2}),d(e^{x-3})$$
and  at the  same time power of  constant  $e$ is increasing corresponding,but my confusion is that  does never  power  in  
$$d(e^{x-c})$$
where  $c$  is some constant  is changed from  -infinite to  +infinity,but  does it never make  $e^{x-c}$  as a constant? or  does never equal $e^{x-c}$   never equal to $1$?  meaning that  $x=c$? if ti makes constant ,then we know that derivative of constant is  zero and whole  multiplication becomes  zero,which is contradiction what  $$d(e^x)=e^x$$
sorry if my idea seems  stupid,but i am curious in this topic and please help me to clarify everything 
 A: When we write $f(x) = e^x$ the $x$ is a variable, it does not have a prescribed value.  So $e^{x-c}$ is not a constant.
It is true that there is a particular value of $x$ for which $e^{x - c} = 1$ because if you pick, for example, $c = 3$ and look at the graph of $e^{x-3}$ it does intersect the horizontal line $y = 1$ (it happens when $x = 3$).  But that's only one value of $x$.  The graph of $e^{x-c}$ is not a horizontal line so the function $e^{x-c}$ is not constant.
The thing to remember here is that the derivative of a function does not depend on the value of that function at a single point.  Instead it depends on how the function behaves around that point.  So even though for every $x$ we could pick a $c$ such that $e^{x - c} = 1$, without changing that $c$ it still would not be the case that $e^{x-c}$ is constant around that particular $x$ so it's derivative will still not be $0$.
Also, remember you can't really say $c$ is infinity because infinity is not a real number that you can plug into an equation.  You can take limits as values tend towards infinity but that's not the same thing as plugging it in.
A: Hmmmmmmmmmm well
\begin{eqnarray*}
d(e^x)&=&d(e\cdot e^{x-1})\\
&=&d(e)\cdot e^{x-1}+e\cdot d(e^{x-1})\\
&=&0\cdot e^{x-1}+e\cdot d(e^{x-1})\\
&=&e\cdot d(e^{x-1})\\
&=&e\cdot (e^{x-1})\\
&=&e^x.
\end{eqnarray*}
