Note that $$\lim_{h \rightarrow 0} (1 +h)^{1/h} = e$$
Let $f(x) = e^x$. Then $f'(x)$ is given by,
$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\lim_{h \rightarrow 0} \frac{e^x(e^h-1)}{h} = \lim_{h \rightarrow 0} e^x\cdot \frac{(e^h-1)}{h} $$
Now since $$e = \lim_{h \rightarrow 0} (1 +h)^{1/h},$$ we have $$\lim_{h \rightarrow 0} e^x\cdot \frac{((\lim_{h \rightarrow 0} (1 +h)^{1/h})^h-1)}{h} =e^x\cdot \lim_{h \rightarrow 0} \frac{((\lim_{h \rightarrow 0} (1 +h)^{h/h})-1)}{h} = \\ e^x \cdot \lim_{h \rightarrow 0} \frac{((\lim_{h \rightarrow 0} (1 +h))-1)}{h} = e^x \cdot \lim_{h \rightarrow 0} \frac{((\lim_{h \rightarrow 0} (h))+1-1)}{h} = \\e^x \cdot \lim_{h \rightarrow 0} \frac{\lim_{h \rightarrow 0} h}{h} = e^x \cdot 1 = e^x$$