This is too long for a comment
- From this you will see that any function that satisfies
$$
f(x + y) = f(x)f(y)
$$
implies that
$$
f(x) = e^{\mu x}
$$
- Call $f(y) = P(X > y)$ in the expression above, we must then have something like
$$
P(X > y) = e^{\mu y}
$$
But since we require that this number vanishes as we approach infinity, this implies that $\mu$ must be negative, $\mu = -\lambda$, for $\lambda > 0$
$$
P(X > y) = e^{-\lambda y} = \int_y^{\infty}{\rm d}y' f_X(y')
$$
- Applying the fundamental theorem of calculus you get
$$
f_X(x) = \lambda e^{-\lambda x}
$$