I've been told that if $$T(f)(x)=g(x)+c\int_0^xK(x,y)f(y)dy$$ for some $c\in\mathbb{R}$ and $f,g\in C[0,1]$ ($g$ fixed) and $K\in C([0,1]\times[0,1])$, then there exists $n\in\mathbb{N}$ such that $T^n$ is a contraction and therefore $T$ has a unique fixed point. I can't seem to work it out. I have that $$T^n(f)=T^{n-1}(g)+c^n\int_0^{x}\int_0^{x_{2}}\cdots\int_0^{x_{n}}\left(\prod_{i=1}^nK(x_i,x_{i+1})f(x_{n+1})\right)dx_{n+1}\cdots dx_2$$ where $x_1=x$, but I have no way of proving this is eventually a contraction. Any ideas? Thanks a lot in advance.
Edit: Writing everything out, it seems (if I did everything correctly) that $||T^n(f)-T^n(h)||_\infty$ can be bounded above by $c^n||K||_\infty^n||f-h||_\infty$, but this doesn't help in order for it to be a contraction.
Edit2 I think the iteration formula is right now, I made a mistake when typing it originally.