Def 1
$\mathcal{K}$ - Volterra integral operator with kernel $K(x,t)$. Repeated kernel $K_n(x,t)$ is the kernel of $\mathcal{K}^n = \underbrace{\mathcal{K} \cdot \mathcal{K} \cdots \mathcal{K}}_{n\text{ times}}$ operator.
Prove, using mathematical induction, that
$K_n(x,t) = \int\limits_t^xK(x,s)K_{n-1}(s,t)ds$
I can guess how to choose base case but I cannot understand how to do the inductive step.
Update: I guess smth with the help of Nick Terziev, please, verify it:
1) $\mathcal{K}f = \int\limits_a^xK(x,s)f(s)ds$
2) $\mathcal{K}^2f = \mathcal{K}\left(\mathcal{K}f\right) = \int\limits_a^xK(x,t)\left(\int\limits_a^tK(t,s)f(s)ds\right)dt = \bigg[a\leq t\leq s\leq x\bigg]=\\=\int\limits_a^xdt\:f(t)\int\limits_t^xK(x,s)K(s,t)ds = \int\limits_a^xK_2(x,t)f(t)dt$
3) $\mathcal{K}^3f = \mathcal{K}\left(\mathcal{K}^2f\right) = \int\limits_a^xK(x,t)\left(\int\limits_a^tK_2(t,s)f(s)ds\right)dt = \\=\int\limits_a^xdt\:f(t)\int\limits_t^xK(x,s)K_2(s,t)ds = \int\limits_a^xK_3(x,t)f(t)dt$
4) So, the base case is: $$K_n(x,t)=\int\limits_t^xK(x,s)K_{n-1}(s,t)ds$$
5) Let's look on $\mathcal{K}^{n+1}$
$\mathcal{K}^{n+1}f=\mathcal{K}\left(\mathcal{K}^nf\right) = \int\limits_a^xK(x,t)\left(\int\limits_a^tK_n(t,s)f(s)ds\right)dt = \\=\int\limits_a^xdt\:f(t)\int\limits_t^xK(x,s)K_n(s,t)ds = \int\limits_a^xK_{n+1}(x,t)f(t)dt$
So, we got $$K_{n+1}(x,t)=\int\limits_t^xK(x,s)K_n(s,t)ds$$
Does it working proof?
You have a mistake in your statement. The right one is that $$K_n(x,t) = \int_t^x ds K(x,s) K_{n-1}(s, t)$$ We will use a change of the order of integration. Indeed, $$\int_a^x K_n(x, t) f(t) dt =\int_a^x K_n(x, s) f(s) ds = \int_a^xK(x, s) ds\int_a^s K_{n-1}(s,t)f(t)dt = \int_a^xdt f(t)\int_t^xds K(x,s)K_{n-1}(s,t)$$ Now we shall compare the last identity with the first and say that as far as it holds for all $f(t)$ your statement is true.