Transformation of the functional equation $f(x+y)=f(x+1)f(y)$

Question

Is there a way to reduce the following functional equation

$$ f(x+y)=f(x+1)f(y),\qquad x,y>0, $$

to the equation

$$ f(x+y)=f(x)f(y),\qquad x,y>0, $$

whose solutions are known?

Thanks in advance.

Answer 1

Let $g(x)=f(x+1)$ then $g(x+y)=f(x+(y+1))=f(x+1)f(y+1)=g(x)g(y)$