Collapsing Infinite Integrals I'm looking into repeated integrals here on Wolfram's Mathworld and I can't seem to figure why it is that the following is true:

\begin{align} \underbrace{\int
 ...\int_0^x}_{n}f\left(x\right)\:\underbrace{dx...dx}_{n}& =
 \int_0^x\frac{f\left(t\right)\left(x-t\right)^{n-1}\:dt}{\Gamma\left(n\right)}.
 \tag{1} \end{align}

I understand the first part, however, s. th. 
\begin{align}
F\left(x\right)=\int f\left(x\right)\:dx=\int_0^x f\left(t\right)\:dt;\:\:F\left(0\right)=0.\tag{2}
\end{align}
It states that
\begin{align}
D^{-n}f\left(x\right)=\underbrace{\int\dots\int_0^x}_{n}f\left(x\right)\underbrace{dx\dots dx}_{n},\tag{3}
\end{align}
and since I know that all of the lower bounds are always $F\left(0\right)=F\left(F\left(0\right)\right)=\dots=0$, that part disappears -- which makes sense. What I don't understand is the collection of the upper bounds. 
After the first integration then, I would have
\begin{align}
\underbrace{\int\dots\int_0^x}_{n-1}F\left(x\right)\:\underbrace{dx\dots dx}_{n-1}=\underbrace{\int\dots\int_0^x}_{n-2}\left(\int_0^x F\left(x\right)\:dx\right)\:\underbrace{dx\dots dx}_{n-2},\tag{4}
\end{align}
But what does the interior of the parenthesis equate to? Surely it is not $F\left(F\left(x\right)\right)$. How do they arrive at $\left(1\right)$?
Thank you for your time,

Edit:
Let $n=1$,
\begin{align}
\int_0^x f\left(x\right)\:dx & = F\left(x\right)-F\left(0\right)=F\left(x\right).\tag{5}
\end{align}
Now let $n=2$, therefore
\begin{align}
\int_0^x\int_0^x f\left(x\right)\:dx\:dx & = \int_0^x\left[F\left(x\right)-F\left(0\right)\right]\:dx=\int_0^x F\left(x\right)\:dx\tag{6}
\end{align}
 A: It looks like you're trying to show that if $F(x) := \int_0^x f(t)\, dt$, then $$F^n(x) = \frac{1}{\Gamma(n)} \int_0^x (x - t)^{n-1}f(t)\, dt,$$
where $F^n$ is $\underbrace{F\circ F \circ \cdots \circ F}_{\text{$n$ times}}$. This can be proven by induction on $n$. 
Since $\Gamma(1) = 1$, the statement when $n = 1$ is $F(x) = \int_0^x f(t)\, dt$, which holds by definition. Consider the $n = 2$ case. We have $$F^2(x) = F\left(\int_0^x f(t)\, dt\right) = \int_0^x \int_0^u f(t)\, dt\, du = \int_0^x \int_t^x f(t)\, du\, dt = \int_0^x (x - t)f(t)\, dt = \frac{1}{\Gamma(2)}\int_0^x (x - t)f(t)\, dt.$$ Now assume $n > 2$ and the result is true for $n$. Then 
\begin{align}F^{n+1}(x) &= F(F^n(x)) = \int_0^x F^n(t)\, dt \\
&= \int_0^x \frac{1}{\Gamma(n)}\int_0^t (t - u)^{n-1} f(u)\, du\, dt \\
&= \frac{1}{\Gamma(n)}\int_0^x \int_u^x (t - u)^{n-1}\, dt\, du\\
&= \frac{1}{\Gamma(n)} \int_0^x \frac{(x - u)^n}{n} f(u)\, du\\
&= \frac{1}{n\Gamma(n)} \int_0^x (x - u)^n f(u)\, du\\
&= \frac{1}{\Gamma(n+1)} \int_0^x (x - u)^n f(u)\, du.
\end{align}
This shows that the result holds for $n+1$.
Note. $\Gamma(n) = (n-1)!$ for all positive integers $n$. So you can write
$$F^n(x) = \frac{1}{(n-1)!} \int_0^x (x - t)^{n-1} f(t)\, dt.$$
