To show that $f(x) =Ae^{nx}$ for constant $n$ and $A$ starting with this thing:
$f'(x) +f(x)=cf(x-1)$
Where $c$ is constant and $c\not= 0$.
If it wasn't for the $f(x-1)$ bit, I would just use the integrating factor where $I=e^x$ and plug it into the equation. But the $f(x)$ throws me off, so I would have to put it into a form like $\frac{dy}{dx}+y=?$, in order to feel ok about it.
EDIT: Oh as somebody rightly pointed out this is only for the condition when $n$ satistfies $n+1 = ce^{-n}$