I'm trying to solve the next problem: Let $f$ be continuous on $\left[0,a\right]$. For $x\in\left[0,a\right]$ define $f_{0}\left(x\right)=f\left(x\right)$ and $f_{n+1}\left(x\right)=\frac{1}{n!}\int_{0}^{x}\left(x-t\right)^{n}f\left(t\right)dt$ for $n\in\mathbb{N}$. Prove that $n$th derivative of $f_{n}$ exists and is equal to $f$.
Using the Leibniz's integral rule is easy to see that $f_{n}'=f_{n-1}$ and with this we obtain that $f_{n}^{\left(n\right)}=f$ for $n\in\mathbb{N}$. But for this rule are used partial derivatives so I think that may be I can't use that, because we haven't seen this in Riemann integral. I also tried to do it using integration by parts. Since $f$ is continuous then $f$ has a primitve $F\left(x\right)=\int_{0}^{x}f\left(t\right)dt$. Then integrating by parts I obtained that
$$f_{n+1}\left(x\right)=\frac{1}{\left(n-1\right)!}\int_{0}^{x}\left(x-t\right)^{n-1}F\left(t\right)dt. $$
The last equality seems some related with $f_{n}$ but I don't know how to differentiate this expression using only fundamental theorem of calculus or I don't know if there is another way to do the problem. Could you help me?
Thanks!
Proceed by induction on $n$. For the inductive step, applying the inductive hypothesis to your result from integration by parts, we get $f_{n+1}^{(n)}(x) = F(x)$, so $f_{n+1}^{(n+1)}(x) = F'(x) = f(x)$.