Nested sum of units

Note that $\begin{aligned} \large \sum_{a_1=0}^4 \sum_{a_2=0}^{a_1} \cdots \sum_{a_n=0}^{a_{n-1}} 1 &= \large \sum_{0 \leq a_n \leq a_{n-1} \leq \cdots \leq a_1 \leq 4} 1 \\ &= \Bigg|\{(a_1,a_2,\ldots,a_n) \in \mathbb{Z}^n : 0 \leq a_n \leq \cdots \leq a_2 \leq a_1 \leq 4\}\Bigg| \\ &= \binom{n+5-1}{5-1} & \text{(Using Stars and Bars)}\\ &= \binom{n+4}{4} \end{aligned}$

Then $\binom{n+4}{4} = 1001 \implies n=\boxed{10}$

Remark : It looks from the comments that there's some confusion about how to use stars-and-bars to count this. Here's an illustration I've given to try to explain it: (note that all "left" and "right" directions are from an outside, fixed observer)

Put $n+4$ people standing side-by-side in a line. Give four of them chairs and have them sit down. Then tell the remaining $n$ people to count how many people to their right are sitting. The number counted by the $k$ th-from-the-left standing person is $a_k.$

This defines a bijection between the number of ways to choose $4$ people from $n+4$ and the number of [non-strictly] decreasing sequences of length $n$ from $\{0,1,2,3,4\}$

$\sum_{a_1=0}^4 1=1+1+1+1+1=5=\binom{5}{4}$ $\sum_{a_1=0}^4\sum_{a_2=0}^{a_1} 1=1+2+3+4+5=15=\binom{6}{4}$ $\sum_{a_1=0}^4\sum_{a_2=0}^{a_1}\sum_{a_3=0}^{a_2} 1=1+3+6+10+15=35=\binom{7}{4}$

$\vdots$ $\sum_{a_1=0}^4 \sum_{a_2=0}^{a_1} \cdots \sum_{a_n=0}^{a_{n-1}} 1 = \binom{4+n}{4}=1001$ So, $n=10$

Proof: $\sum_{a_1=0}^m 1=m+1=\binom{m+1}{1}$

$\sum_{a_1=0}^m\sum_{a_2=0}^{a_1} 1=\sum_{a_1=0}^m \binom{a_1+1}{1}=\binom{m+2}{2}$

$\sum_{a_1=0}^m\sum_{a_2=0}^{a_1}\sum_{a_3=0}^{a_2} 1=\sum_{a_1=0}^m \binom{a_1+2}{2}=\binom{m+3}{3}$

$\sum_{a_1=0}^4\sum_{a_2=0}^{a_1}\sum_{a_3=0}^{a_2}\sum_{a_4=0}^{a_3} 1=\sum_{a_1=0}^{m}\binom{a_1+3}{3}=\binom{m+4}{4}$

$\vdots$

$\sum_{a_1=0}^m \sum_{a_2=0}^{a_1} \cdots \sum_{a_n=0}^{a_{n-1}} 1=\sum_{a_1=0}^m\binom{a_1+(n-1)}{n-1}=\binom{m+n}{n}$

Put in $m=4$ will get the above result.

I needed to reason this out without all those pesky Sigmas.

If n=0 it's just a single 1 with no summation. $1$

If n=1 it's just the number 1 added to itself 5 times. $1+1+1+1+1=5$

If n=2 we need to make the next summation do each of 0,1,2,3,4 and so the sum is $(1)+(1+1)+(1+1+1)+(1+1+1+1)+(1+1+1+1+1)=1+2+3+4+5=15$

If n=3 we need to make the next summation do each of 0,1,2,3,4 on the previous so the sum is $(1)+(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)=1+3+6+10+15=35$

Adding numbers like this is one way of using Pascal's Triangle: the Hockey Stick Identity

We want hockey sticks 5 numbers long. This means we want the 5th number in each row (counting starts with 0)

A glance at the Triangle shows $\binom{14}{4}=1001$ but since n=1 is in row 5: 4 numbers ahead, the solution is $n=14-4=\boxed{10}$

Define $\displaystyle{f(x,n) = \sum_{a_{1} = 0}^{x}\sum_{a_{2} = 0}^{a_{1}}\cdots\sum_{a_{n} = 0}^{a_{n-1}}1}$ . Immediately it follows that $f(0,n) = 1$ and $f(x,n) = 0$ for all $x < 0$ . Moreover, we have: $f(x,n) = f(x-1,n)+f(x,n-1).$ To keep the above relation consistent, set $f(x, 0) = 1$ . Consider this sequence: $\begin{aligned} {\color{#3D99F6} 1}f(x&,n)= \\ {\color{#3D99F6} 1}f(x-1,n)\, &+\, {\color{#3D99F6} 1}f(x,n-1)= \\ {\color{#3D99F6} 1}f(x-2,n)\, +\, {\color{#3D99F6} 2}f(x-1&,n-1)\, +\, {\color{#3D99F6} 1}f(x,n-2)= \\ {\color{#3D99F6} 1}f(x-3,n)\, +\, {\color{#3D99F6} 3}f(x-2,n-1)\, &+\, {\color{#3D99F6} 3}f(x-1,n-2)\, +\, {\color{#3D99F6} 1}f(x,n-3) = \\ &~\vdots\end{aligned}$ One can easily recognize that coefficients correspond to Pascal triangle. Next, notice that coefficient of $f(x-a, n-b)$ is equal to $\displaystyle{\binom{a+b}{a} = \binom{a+b}{b}}$ .

Further on, deduce that the whole sum is going to converge to one element of our Pascal triangle, namely the one corresponding to $f(0,0)$ : $f(x, n) = \binom{x + n}{x} \cdot f(0,0) = \binom{x+n}{x} \cdot 1$

Finally, plugging in values $x = 4$ and $f(4, n) = 1001$ yields $n = 10.$

Finish 9 sums at the right end of the expression and simplify it as:

[(j+9)!/{(j!)(9!)}].

After 1 more explicit summation and using WolframAlpha get:

[sum_(j=0,4){((j + 9)!)/(9!)(j!)}] = 1001

Answer=10