Differentiability of $f(x+y) = f(x)f(y)$ Let $f$: $\mathbb R$ $\to$ $\mathbb R$ be a function such that $f(x+y)$ = $f(x)f(y)$ for all $x,y$ $\in$ $\mathbb R$. Suppose that $f'(0)$ exists. Prove that $f$ is a differentiable function.
This is what I've tried:
Using the definition of differentiability and taking arbitrary $x_0$ $\in$ $\mathbb R$.
$\lim_{h\to 0}$ ${f(x_0 + h)-f(x_0)\over h}$ $=$ $\cdots$ $=$  $f(x_0)$$\lim_{h\to 0}$ ${f(h) - 1\over h}$.
Then since $x_0$ arbitrary, using $f(x_0+0) = f(x_0) = f(x_0)f(0)$ for $y = 0$, can I finish the proof?

By assumption 
$$
\lim_{h \to 0}\frac{f(h) - f(0)}{h} = f(0)\lim_{h \to 0}\frac{f(h) - 1}{h}
$$
exists,
implying 
$$
l := \lim_{h \to 0}\frac{f(h)-1}{h}
$$
exists;
hence,
if $x \in \Bbb{R}$, then
$$
\frac{f(x+h) - f(x)}{h} = \frac{f(x)[f(h) - 1]}{h} \to f(x)l
$$
as $h \to 0$.
