Prove that $\sum_{{a_{n-1}}=1}^{a_n}\,\sum_{a_{n-2}=1}^{a_{n-1}}\,\ldots\,\sum_{a_{1}=1}^{a_2}\,a_1=\frac{\prod\limits_{i=0}^{n-1}\,(a_n+i)}{n!}$. 
Use Principle of Mathematical Induction to show that, for every integer $n\ge2$,
  $$\sum_{{a_{n-1}}=1}^{a_n}\,\sum_{a_{n-2}=1}^{a_{n-1}}\,\ldots\,\sum_{a_{1}=1}^{a_2}\,a_1=\frac{\prod\limits_{i=0}^{n-1}\,(a_n+i)}{n!}\,.$$

I have totally no idea to deal with the product. I can understand it when it only uses 3 sigma notation.   
 A: The correct statement is $$\sum_{{a_{n-1}}=1}^{a_n}\,\sum_{a_{n-2}=1}^{a_{n-1}}\,\ldots\,\sum_{a_{1}=1}^{a_2}\,a_1=\frac{\prod\limits_{i=0}^{n-1}\,(a_n+i)}{n!}\text{ for all }n\in\mathbb{Z}_{>0}\text{ and }a_n\in\mathbb{Z}_{\geq 0}\,.$$  This is equivalent to saying that
$$\sum_{{a_{n-1}}=1}^{m}\,\sum_{a_{n-2}=1}^{a_{n-1}}\,\ldots\,\sum_{a_{1}=1}^{a_2}\,a_1=\binom{m+n-1}{n}\text{ for all }m\in\mathbb{Z}_{\geq 0}\text{ and }n\in\mathbb{Z}_{>0}\,.\tag{*}$$ 
To prove by induction, you can justify the Hockey-Stick Identity $$\sum_{k=1}^N\,\binom{k}{r}=\binom{N+1}{r+1}\text{ for all }N,r\in\mathbb{Z}_{\geq 0}\,.\tag{#}$$
(Therefore, if you want to prove everything from the Hockey-Stick Identity to your own identity, then you need two induction proofs.  First, prove (#) by induction on $N$.  Then prove (*) by induction on $n$.)
Now that I have given a hint on how to do the job inductively, I present a combinatorial approach.  Rewrite (*) as
$$\sum_{{a_{n-1}}=1}^{m}\,\sum_{a_{n-2}=1}^{a_{n-1}}\,\ldots\,\sum_{a_{1}=1}^{a_2}\,\sum_{a_0=1}^{a_1}\,1=\binom{m+n-1}{n}\text{ for all }m\in\mathbb{Z}_{\geq 0}\text{ and }n\in\mathbb{Z}_{>0}\,.$$
The left-hand side of the equation above counts the number of $n$-tuples $\left(a_0,a_1,a_2,\ldots,a_{n-1}\right)$ of integers $a_0,a_1,a_2,\ldots,a_{n-1}$ such that $$1=:a_{-1}\leq a_0 \leq a_1 \leq a_2\leq \ldots \leq a_{n-1}\leq a_n:=m\,.$$
Let $x_j:=a_j-a_{j-1}$ for $j=0,1,2,\ldots,n$.  Then, $(x_0,x_1,x_2,\ldots,x_n)$ is an $(n+1)$-tuple of nonnegative integers such that $$x_0+x_1+x_2+\ldots+x_n=m-1\,.$$ 
(We can get $a_0,a_1,a_2,\ldots,a_{n-1}$ back by noting that $a_j=1+x_0+x_1+\ldots+x_j$ for every $j=0,1,2,\ldots,n-1$.
By the stars-and-bars technique, there are precisely
$$\binom{(m-1)+(n+1)-1}{(n+1)-1}=\binom{m+n-1}{n}$$ solutions $(x_0,x_1,\ldots,x_n)$.  This proves (*).
