I am having difficulty proving the final bit of Campbell's formula, that is, given $C_t = H_1 + \cdots + H_{N_t}$ a compound Poisson process with iid jumps $H_k ~ \mu$ and an independent Poisson process $N_t$ with intensity $\lambda$, we have $$E \exp (i\int_0^\infty f(t+s)dC_t) = \exp (\lambda \int_0^\infty \int_{y\neq 0} (e^{iyf(s+t)}-1) \mu(dy)dt)$$ holds for all $s\ge 0$ and bounded measurable $f:[0,\infty) \to \mathbb{R}^d$ with compact support.
The proof goes by defining $\phi(s):=E\exp(i\int_0^\infty f(s+t)dC_t)$. And then $\gamma(s+x):= E\exp(if(s+x)H_1)$.
Then using iid, we can get $$e^-\lambda s\phi(s)=\lambda \int_s^\infty (\phi(t)e^{-\lambda t})\gamma(t)dt.$$ Since $f$ has compact support $\phi(\infty)=1$.
Using this integral equation, how do we show that the right hand side of the first displayed equation is the unique solution?