How to prove $\dfrac{d}{dx}e^x=e^x$? I have a problem : How to prove
$$\dfrac{d}{dx}e^x=e^x\;?$$
My answer is
\begin{eqnarray}
\dfrac{d}{dx}e^x&=&\lim\limits_{h\to 0}\dfrac{e^{x+h}-e^x}{h}\\
&=&\lim\limits_{h\to 0}\dfrac{e^{x}e^h-e^x}{h}\\
&=&e^{x}\lim\limits_{h\to 0}\dfrac{e^h-1}{h}\\
\end{eqnarray}
But I don't know to find $\lim\limits_{h\to 0}\dfrac{e^h-1}{h}$. So how to find it?
Note:  I cannot use L'Hospital rule.
 A: The key fact is that
$$\lim_{n\to \infty} \left(1+\frac1n\right)^n \to e \implies \lim_{n\to \infty} \left(1+\frac1x\right)^x \to e$$
indeed for any $x\in (n,n+1)$ 
$$\left(1+\frac1{n+1}\right)^n \le \left(1+\frac1x\right)^x \le \left(1+\frac1{n}\right)^{n+1} $$
then take the limit and use squeeze theorem.
Then
$$\lim_{x\to \infty} \left(1+\frac1x\right)^x \to e \implies \frac{\log \left(1+\frac1x\right)}{\frac1x}\to 1 \implies \frac{\log (1+t)}{t}\to 1 \, t\to 0$$
then by $h=\log (1+t)\to 0,  t \to 0$
$$\lim\limits_{h\to 0}\dfrac{e^h-1}{h}=\lim\limits_{t\to 0}\dfrac{t}{\log (1+t)}=1$$
A: Personally, the simplest and best way to prove this identity is not from the definition of derivative, but to use the power series for $e^x$:
$$e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{2 \cdot 3} + ...$$
Since
$$\frac{d}{dx} \left( f(x) + g(x) \right) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$$
performing the derivative becomes quite easy and should easily show the identity.
The power series will also prove useful if going from the definition of derivative as a limit as well, if you insist on using that. Personally, the method above I find much more elegant and intuitive but that's a matter of opinion.

Note that interchanging the operations of derivatives and infinite summation does not always hold, i.e.
$$\frac{d}{dx} \sum_{k=0}^\infty f(k) = \sum_{k=0}^\infty \frac{d}{dx} f(k)$$
need not always hold. (While the derivative operation does "distribute" per the summation rule above, that is for finite summation, and need not hold true when we have infinite summation. That it does is a consequence of other facts.)
My own personal understanding of the underlying reason is lacking, so I'm mostly deferring to Riley's comments on my answer in the matter. For this to be applicable in general, the sequence of derivatives for $f(k)$ must converge uniformly - which, in this case, happens to be true for the polynomials $f(k) = x^k/k!$.
Edit
The proof that a power series with uniform convergence may be differentiated termwise within its radius of convergence is given as theorem 2 on this page and on this page. The proof that $e^x$ has a radius convergence of $\infty$ (and hence is termwise differentiable everywhere) is given on this page.
A: \begin{eqnarray}
e^x&=&\sum\limits_{k=0}^\infty \dfrac{x^k}{k!}\\
\dfrac{d}{dx}e^x&=&\dfrac{d}{dx}\sum\limits_{k=0}^\infty \dfrac{x^k}{k!}\\
&=& \sum\limits_{k=0}^\infty \dfrac{d}{dx}\dfrac{x^k}{k!}\\
&=& \sum\limits_{k=1}^\infty \dfrac{kx^{k-1}}{k!}\\
&=& \sum\limits_{k=1}^\infty \dfrac{x^{k-1}}{(k-1)!}\\
&=& \sum\limits_{k=0}^\infty \dfrac{x^{k}}{k!}\\
&=&e^x
\end{eqnarray}
