Answer
See the proof.
Work Step by Step
We know that:
$$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$
$$f'(x)=\lim_{h \to 0}\frac{f(x)f(h)-f(x)}{h}$$
$$f'(x)=f(x)\lim_{h \to 0}\frac{f(h)-1}{h}$$
Since $\displaystyle \lim_{h \to 0}\frac{f(h)-1}{h}=\frac{0}{0}$ because $f(0)=1$ then by the l'Hospital's rule it follows:
$$\displaystyle \lim_{h \to 0}\frac{f(h)-1}{h}=\displaystyle \lim_{h \to 0}\frac{f'(h)}{1}=\displaystyle \lim_{h \to 0}f'(h)=f'(0)=1$$
so:
$$f'(x)=f(x)\lim_{h \to 0}\frac{f(h)-1}{h} \to f'(x)=f(x)\cdot 1=f(x)$$
---------------------------------------------------------------------
Since:
$$f'(x)=f(x)$$
$$\frac{f'(x)}{f(x)}=1$$
$$\int \frac{f'(x)}{f(x)}dx=\int 1dx$$
$$\ln(|f(x)|)=x+c$$
Since $f(0)=1$ then $\ln(|f(0)|)=0+c \to \ln(|1|)=c \to 0=c$
so:
$$\ln(|f(x)|)=x+0=x$$
$$\ln(|f(x)|)=x$$
$$|f(x)|=e^{x}$$
$$f(x)=e^{x}~~\text{or}~~f(x)=-e^{x}$$
Since $f(0)=1$ so the solution $f(x)=-e^{x}$ should be discarded because $f(0)=-e^{0}=-1 \neq 1$.
so the solution is:
$$f(x)=e^{x}$$
