N-th derivative of n-fold integral I want to justify that the n-th derivative of an n-fold integral gives the original function. In other words that
$$
\frac{d^n}{dx^n}\frac{1}{(n-1)!}\int_{0}^{x}(x-s)^{n-1}f(s)ds=f(x)
$$
If I substitute n=1, then is clear that the equation holds. But
If I substitute n=2, then:
$$
\frac{d^2}{dx^2}\frac{1}{(2-1)!}\int_{0}^{x}(x-s)^{2-1}f(s)ds=f(x)
$$
which is:
$$
\frac{d^2}{dx^2}\int_{0}^{x}(x-s)f(s)ds=f(x)
$$
Now I can split this like this:
$$
\frac{d^2}{dx^2}\int_{0}^{x}xf(s)ds-
\frac{d^2}{dx^2}\int_{0}^{x}sf(s)ds
$$
Which wolfram alpha says that it is:
$$
xf'(x)+2f(x)-(xf'(x)+f(x))
$$
That clearly gives f(x). But how do I prove that the second derivatives of such integrals in fact gives those results?
 A: I think I've got the answer, using the mathematical induction. We have verify that the formula is valid for n=1 and n=2, also n=3 (but not shown here).
Now we make our hypothesis:
I.H. : 
$$
\frac{1}{(n-1)!}\frac{d^n}{dx^n}\int_{0}^{x}(x-s)^{n-1}f(s)ds=f(x)
$$
Now we make the inductive step, which is that for n+1 we will got f(x) also, so make n+1 in the above formula:
$$
\frac{1}{(n)!}\frac{d^{n+1}}{dx^{n+1}}\int_{0}^{x}(x-s)^{n}f(s)ds=f(x)
$$
Which is equal to:
$$
\frac{1}{(n)!}\frac{d^n}{dx^n}(\frac{d}{dx}\int_{0}^{x}(x-s)^{n}f(s)ds)=
$$
Using the Leibniz Rule in its integral form we have:
$$
\frac{1}{(n)!}\frac{d^n}{dx^n}((x-x)^{n}f(x)+\int_{0}^{x}n(x-s)^{n-1}f(s)ds)=
$$
$$
\frac{n}{(n)!}\frac{d^n}{dx^n}(\int_{0}^{x}(x-s)^{n-1}f(s)ds)=
$$
$$
\frac{n}{1*2*3...*(n-1)*n}\frac{d^n}{dx^n}(\int_{0}^{x}(x-s)^{n-1}f(s)ds)=
$$
$$
\frac{1}{(n-1)!}\frac{d^n}{dx^n}(\int_{0}^{x}(x-s)^{n-1}f(s)ds)
$$
Using the I.H. we observe that this is equal to f(x):
$$
\frac{1}{(n-1)!}\frac{d^n}{dx^n}(\int_{0}^{x}(x-s)^{n-1}f(s)ds)=f(x)
\\
\square
$$
A: I think I've got the answer. We have to use the Leibniz's rule integral form.
So for example 
$$
\frac{d^2}{dx^2}\int_{0}^{x}xf(s)ds=\frac{d}{dx}(xf(x)+\int_{0}^{x}f(s)ds)
$$
which is equal to:
$$
=xf'(x)+f(x)+f(x)+\int_{0}^{x}0ds=xf'(x)+2f(x)
$$
As wolfram alpha said.
