Calculating the $p+1$ derivative I'm stucked with the following:

Let $y\in C^1(I),f\in C^p(I\times R,R)$,$k\in C^p(I\times I,R)$, $f$ is Lipschitz and such that
  $$y'(t)=f(t,y(t))+g(t)$$
  where
  $$g(t)= \int_{t_0}^t k(t,s)y(s)\mathrm{d}s\qquad y(t_0)=y_0$$
  Prove that $y\in C^{p+1}$

I see that I only have to prove that $\eta\in C^p$ when
 $$\eta(t):=\int_{t_0}^t k(t,s)y(s)\mathrm{d}s$$
I tried to prove it by induction, but I fail to see what I should assume to be true for the $p=n$, since I don't know the formula for the $n$-derivative.
If $p=1$
\begin{align*}
\eta^{p}(t)&=k(t,t)y(t) +\int_{t_0}^t \frac{\mathrm{d}}{\mathrm{d}t} k(t,s)y(s)\mathrm{d}s\\
&=k(t,t)y(t)+\int_{t_0}^t k_t(t,s)y(s)\mathrm{d}s
\end{align*}
Now what should I do now? If anyone could give me a hint maybe?
 A: We have $y'(t)=f(t,y(t))+\int_{t_0}^t k(t,s)y(s)\mathrm{d}s$ and the initial hypothesis of $y'$, $f^{p}$, $k_{x^p}$ and $k_{y^p}$ being continuous. We can prove that $y''$ is continuous too.
$$k_1(u,v)=k(u,v)$$
$$y''(t)=\frac{\mathrm d}{\mathrm dt}f(t,y(t))+k_1(t,t)y(t) +\int_{t_0}^t \frac{\partial}{\partial t} k_1(t,s)y(s)\mathrm{d}s$$
Is continuos because $\dfrac{\mathrm d}{\mathrm dt}f(t,y(t))$ is the derivative of a composition of functions with continuous derivatives and $k_1$ has partial derivatives continuous.
$$k_2(u,v)=\frac{\partial}{\partial u} k_1(u,v)$$
$$y'''(t)=\frac{\mathrm d^2}{\mathrm dt^2}f(t,y(t))+\frac{\mathrm d}{\mathrm dt}(k_1(t,t)y(t)) +k_2(t,t)y(t)+\int_{t_0}^t \frac{\partial}{\partial t} k_2(t,s)y(s)\mathrm{d}s$$
Is continuous because is the sum of functions or its derivatives of functions with at least its second derivatives continuous.
Now, define,
$$k_n(u,v)=\frac{\partial}{\partial u} k_{n-1}(u,v)$$
Consider this formula for the $(n+1)$-th derivative of $y$
$$y^{(n+1)}(t)=\frac{\mathrm d^n}{\mathrm dt^n}f(t,y(t))+\sum_2^{n}\frac{\mathrm d^{i-1}}{\mathrm dt^{i-1}}(k_{n-i+1}(t,t)y(t)) + k_n(t,t)y(t)+\int_{t_0}^t \frac{\partial}{\partial t} k_{n}(t,s)y(s)\mathrm{d}s$$
Differentiate once more time:
$$y^{(n+2)}(t)=\frac{\mathrm d^{n+1}}{\mathrm dt^{n+1}}f(t,y(t))+\sum_2^{n}\frac{\mathrm d^{i}}{\mathrm dt^{i}}(k_{n-i+1}(t,t)y(t)) + \frac{\mathrm d}{\mathrm dt}k_n(t,t)y(t)+k_{n+1}(t,t)y(t)+\int_{t_0}^t \frac{\partial}{\partial t} k_{n+1}(t,s)y(s)\mathrm{d}s$$
$$y^{(n+2)}(t)=\frac{\mathrm d^{n+1}}{\mathrm dt^{n+1}}f(t,y(t))+\sum_2^{n+1}\frac{\mathrm d^{i-1}}{\mathrm dt^{i-1}}(k_{n+1-i+1}(t,t)y(t))+k_{n+1}(t,t)y(t)+\int_{t_0}^t \frac{\partial}{\partial t} k_{n+1}(t,s)y(s)\mathrm{d}s$$
So, the formula holds for $n+1$ if it holds for $n$ and, too, if $y$ has its $n+1\le p$ derivative continuous, it has derivative $n+2$ continuous. 
Now, set $n+1=p$ and it's proved.
