analitycs solutions to the equation $f'(x)=f(x)f(x-1)$ As the title says I'm serching for functions ($C^n$ or analitycs $f$) that satisfies 
$f'(x)=f(x)f(x-1)$

some details: I've come at this equation after looking for a function $g$ satisfying for some given $h$ 
$g'(g^{-1}(x))=x\cdot h(x)$
I tried to manipulate the equation 
$g'(g^{-1}(g(x)))=g(x)\cdot h(g(x))$
And since I assume (define)that $h(g(x))=g(x-1)$ holds the equation should turn into
$g'(x)=g(x)g(x-1)$

Define $f(x)$ for $x\in[0,1]$ to be anything.
$$\frac{f'(x)}{f(x)}=f(x-1)\\
\int\frac{f'(x)}{f(x)}dx=\int f(x-1)dx\\
\ln(f(x))+c=\int f(x-1) dx\\
f(x)=A\exp(\int f(x-1) dx)$$
Then, since you already know $f(1)$, 
$$f(x)=f(1)\exp(\int_1^x f(t-1)dt)$$ 
defines $f(x)$ for $x\in[1,2]$
