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 $x\in[1,2]$