$(e^x)'=e^x$

$f(x)=e^x$;

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{e^{x+h}-e^x}{h}=\displaystyle\lim_{h\longrightarrow 0} \dfrac{e^x(e^h-1)}{h} = e^x\lim_{h\longrightarrow 0} \dfrac{(e^h-1)}{h}=e^x$

where in the last limit we have used the derivative of the exponential in the origin

$\displaystyle\lim_{h\longrightarrow 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\longrightarrow 0}\dfrac{e^{x+h}-e^x}{h}=\displaystyle\lim_{h\longrightarrow 0} \dfrac{e^x(e^h-1)}{h} = e^x\lim_{h\longrightarrow 0} \dfrac{(e^h-1)}{h}=e^x$

where in the last limit we have used the derivative of the exponential in the origin