proof that $\frac{d}{dx}e^x = e^x$ I am working on proving different mathematical formulas. I am currently working on proving that $\frac{d}{dx} e^x = e^x$ . This is my proof so far:
$$\frac{d}{dx} e^x = \lim_{h \to0}\frac{e^{x+h} - e^{x}}{h} = $$
$$ \lim_{h \to 0} \frac{e^x(e^h-1)}{h} = e^x \cdot\lim_{h \to 0} \frac{(e^h-1)}{h}$$
Need to prove that $\lim_{h \to 0} \frac{(e^h-1)}{h} = 1$.
Using substitution: $h = \ln(t+1)$
$$\lim_{t \to 0}\frac{t}{\ln(t+1)}$$
$$\lim_{t \to 0}\frac{t \cdot \frac{1}{t}}{\ln(t+1)\cdot \frac{1}{t}}$$
$$\lim_{t \to 0}\frac{1}{\ln(t+1)^{\frac{1}{t}}}$$
$$\frac{1}{\ln(e)} = \frac{1}{1} = 1$$
$\blacksquare$

Am I allowed to say $t \to 0$ in a proof when it is actually $\ln(t+1) \to 0$?
Do you have any corrections on the proof?
 A: For exponential you cannot use Taylor series or L'Hopital since you will stuck into recursive reasoning.
Note that exponential is introduced in math after natural logarithm. So use function inverse:
$$y=\exp(x)$$
$$x=\ln(y)$$
$$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{\frac{1}{y}}=y=\exp(x)$$
Additional explanation
Definition of $\ln (x)$:
$$\ln(x)=\int_1^x\left(\frac1t\right)dt$$
And $\exp(x)$ is defined as inverse function of $\ln(x)$.
A: That $e$ is the unique number $a$ for which $\displaystyle\lim_{h\to0}\frac{a^h-1}h$ is equal to $1$ is sometimes taken to be the definition of $e$.
Notice that with $a=1$ we can see that the limit is less than $1$, as follows: As $h$ goes from $0$ to $1$, $a^h=2^h$ goes from $1$ to $2$, so the slope of that secant line is $1$.  Since the tangent line gets steeper as you go from left to right, the slope of the tangent line at $0$ must therefore be less than $1$.  You can also show it's more than $1/2$ by considering the secant line through $h=0$ and $h=-1$.
Using $h=0$ and $h=-1/2$, you can show in the same way that $\displaystyle\lim_{h\to0}\frac{4^h-1}h$ is bigger than $1$.
Thus the number $a$ for which $\displaystyle\lim_{h\to0}\frac{a^h-1}h$ is exactly $1$ must be somewhere between $2$ and $4$.
There are many characterizations of $e$.  Which of them is taken to be the definition depends on context and is a judgment call.
A: A meta-answer: Each of the three answers so far are coming from a different definition of the exponential function.  They are all equivalent.  But in each course, only one of them is the definition.  
So it's important to understand what the definitions are in your course before you can prove this statement yourself.
A: $$\frac{d}{dx}e^x=\frac{d}{dx}\sum_{k=0}^\infty \frac{x^k}{k!}\underset{(why\ ?)}{=}\sum_{k=0}^\infty \frac{d}{dx}\left(\frac{x^k}{k!}\right)=\sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!}\underset{\ell=k-1}{=}\sum_{\ell=0}^\infty \frac{x^\ell}{\ell !}=e^x.$$
