TITLE: Transformation of the functional equation $f(x+y)=f(x+1)f(y)$
QUESTION [6 upvotes]: 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.

REPLY [13 votes]: Let $g(x)=f(x+1)$ then $g(x+y)=f(x+(y+1))=f(x+1)f(y+1)=g(x)g(y)$