Proof of multiple integral equation I am trying to prove the following equation:
$$
\int_a^x\int_a^{t_1}\cdots\int_a^{t_{n-1}}\,dt_n\cdots\,dt_2\,dt_1=\frac{(x-a)^n}{n!}$$
I am not really sure where to begin. I would appreciate any help.
 A: Let $I_n(x)$ be the integral above. It is straightforward to verify the
equation for $n=1$. 
Note that we have $I_{n+1}'(x) = I_n(x)$.
Suppose the equation is true for $1,...,n$.
Then integrating the equation above gives
$I_{n+1}(x) = \int_a^x I_{n+1}'(t)dt = \int_a^x I_{n}(t)dt= \int_a^x {(t-a)^n \over n! }dt $, and evaluating the integral yields the desired result.
A: @Zardo suggested mathematical induction on $n$.  Let's try a crude form of that and see if you can take it from there.
$$
\int_a^{t_{n-1}} \, dt_n = t_{n-1} - a.
$$
\begin{align}
\int_a^{t_{n-2}} \int_a^{t_{n-1}} \, dt_n\,dt_{n-1} & = \int_a^{t_{n-2}} (t_{n-1} - a)\,dt_{n-1} \\[8pt]
& = \left.\vphantom{\frac11} \frac{(t_{n-1} - a)^2} 2 \right|_{t_{n-1}:=a}^{t_{n-1}:=t_{n-2}} = \frac{(t_{n-2}-a)^2} 2
\end{align}
\begin{align}
\int_a^{t_{n-3}} \int_a^{t_{n-2}} \int_a^{t_{n-1}} \, dt_n\,dt_{n-1} \, dt_{n-2} & = \int_a^{t_{n-3}} \frac{(t_{n-2}-a)^2} 2\,dt_{n-1} \\[8pt]
& = \left.\vphantom{\frac11} \frac{(t_{n-2} - a)^3} 6 \right|_{t_{n-2}:=a}^{t_{n-2}:=t_{n-3}} =  \frac{(t_{n-3}-a)^3} 6
\end{align}
$\ldots$ and so on.  That is how mathematical induction is done, but it's stated somewhat more abstractly.
Let's make it a little bit more abstract:
Induction hypothesis:
$$
\int_a^{t_{n-(k-1)}} \int_a^{t_{n-(k-2)}} \cdots \int_1^{t_{n-1}} dt_n \, dt_{n-1} \cdots dt_{n-(k-2)} = \frac{(t_{n-(k-1)}-a)^{n-(k-1)}}{(n-(k-1))!}.
$$
Induction step:
\begin{align}
& \int_a^{t_{n-k}} \int_a^{t_{n-(k-1)}} \int_a^{t_{n-(k-2)}} \cdots \int_a^{t_{n-1}} dt_n \, dt_{n-1} \cdots dt_{n-(k-2)} \, dt_{n-(k-1)} \\[10pt]
= {} & \int_a^{t_{n-k}} \frac{(t_{n-(k-1)}-a)^{n-(k-1)}}{(n-(k-1))!}  \, dt_{n-(k-1)} \\[10pt]
= {} & \left. \frac{(t_{n-(k-1)} - a)^{n-k}}{(n-k)!} \right|_{t_{n-(k-1)}\,:=\,a}^{t_{n-(k-1)}\,:=\,t_{n-k}} \\[10pt]
= {} & \frac{(t_{n-k} -a)^{n-k}}{(n-k)!}.
\end{align}
This is the same argument that was presented more concretely above.
A: HINT:
$$\begin{align}
\int_a^{t_{n-1}}dt_n&=\frac{1}{1}(t_{n-1}-a)^1\\\\
\int_a^{t_{n-2}}(t_{n-1}-a)dt&=\frac{1}{2\cdot 1}(t_{n-2}-a)^2\\\\
\int_a^{t_{n-3}}\frac12 (t_{n-2}-a)^2dt&=\frac{1}{3\cdot 2\cdot 1}(t_{n-3}-a)^3
\end{align}$$

HINT 2:
We have established a benchmark that the result is true if $n=3$ ($x=t_{n-3}=t_0$).  Now assume that the relationship is true for some number $k$.  Show that it is true for $k+1$ to complete the proof by induction.
