We first tackle the simplest case in which $g(x) = x$, then we could use chain rule to determine the result. 
So, for $e^{x}$ using the definition of derivative as the limit ratio,
$$
\lim_{h \to 0} \frac{e^{x+h}-e^{x}}{h}= e^{x} \lim_{h\to 0} \frac{e^h - 1}{h}
$$
where the latter limit is just $1$, one could see it as a consequence of the definition of $e^h = \sum_{n=0}^{+\infty} \frac{h^n}{n!}$. Finally, we have that the derivative of $e^x$ is $e^x$. By applying chain rule to the aforementioned question, we conclude 
$$
\frac{\textrm{d} e^{g(x)}}{\textrm{d}x} = e^{g(x)} \frac{\textrm{d} g(x)}{\textrm{d}x}
$$ 
where the Leibniz notation for the derivatives was used.
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